# Two explicit Transcendental Numbers

In my fresh look at Galois Theory that I’ve had a year ago, I was content when I arrived at the proofs for the impossibility of the classical geometry problems (squaring the circle and the like) and for the Fundamental Theorem of Algebra (which I presented here). A gap in the business about squaring the circle was the proof that $\pi$ is actually not contained in some finite field extension of $\mathbb{Q}$ having a degree of a power of $2$. Actually, as Lindemann proved in 1882, there is a stronger result: $\pi$ is even transcendental. I have taken some time to look at the proof of this, and as a by-product, we can also prove that $e$ is transcendental.

Again, quite a stronger result holds which is known as the Lindemann-Weierstrass-Theorem: If the equation

$\displaystyle \sum_{k=1}^nc_ke^{\alpha_k} = 0$

holds for some $n\in\mathbb{N}$, algebraic numbers $c_k$, and pairwise distinct algebraic numbers $\alpha_k$ – then all the $c_k$ must be equal to $0$. We won’t prove that, even though the ideas are roughly the same as presented here. The results that follow are special cases of this theorem.

By usual cardinality arguments, it is easy to see that there are very many transcendental numbers. Actually putting your finger on one of these numbers is quite hard. Liouville was the first to give a criterion for this in 1851. Hermite proved in 1873 that $e$ is transcendental, and 9 years later, Lindemann followed up about $\pi$.

I have learned these proofs from an exposition I found on the internet, and hence the notation will be very close to this exposition. I also took some clues from a German book for undergraduate students. In this book, the complete Lindemann-Weierstrass-Theorem is proved.

We will first be concerned with the following statement:

Theorem: The number $e=\sum_{k=1}^\infty\frac1{k!}$ is transcendental.

For the proof we will apply the function

$I(t) = \displaystyle\int_0^te^{t-u}f(t)du,\qquad t\in\mathbb{C}.$

Let $f\in\mathbb{C}[x]$ be a polynomial of degree $n$.

If the coefficients of $f$ are $a_k\in\mathbb{C}$, we will use the abbreviation $\overline f(x)$ for the polynomial which has $\left|a_k\right|$ as its coefficients. Then, it is immediate:

$\displaystyle \left|I(t)\right|\leq\int_0^t\left|e^{t-u}f(u)\right|du\leq\left|t\right|\max\{\left|e^{t-u}\right|\}\max\{\left|f(u)\right|\}\leq\left|t\right|e^{\left|t\right|}\overline{f}(\left|t\right|).$

Integration by parts and a little induction will show

The Integral-Lemma: Let $I$ and $f$ be defined as above. Then,

$\displaystyle I(t) = e^t\sum_{j=0}^nf^{(j)}(0)-\sum_{j=0}^nf^{(j)}(t)$.

Proof: Let us integrate $I$ by parts:

\begin{aligned}\displaystyle \int_0^te^{t-u}f(u)du &= \left[-e^{t-u}f(u)\right]_0^t-\int_0^t(-e^{t-u})f'(u)du \\ &\displaystyle= -f(t)+e^{t}f(0)+\int_0^te^{t-u}f'(u)du. \end{aligned}

For the inductive step, let’s assume we have done this integration by parts $k$ times already and do it now once again:

\begin{aligned} \displaystyle \int_0^te^{t-u}f(u)du &= -\sum_{j=0}^k f^{(j)}(t)+e^t\sum_{j=0}^kf^{(j)}(0)+\int_0^te^{t-u}f^{(j+1)}(u)du\\ &=\displaystyle -\sum_{j=0}^k f^{(j)}(t)+e^t\sum_{j=0}^kf^{(j)}(0)+ \left[-e^{t-u}f^{(j+1)}(u)\right]_0^t-\int_0^t(-e^{t-u})f^{(j+2)}(u)du\\ &=\displaystyle -\sum_{j=0}^k f^{(j)}(t)+e^t\sum_{j=0}^kf^{(j)}(0)- f^{(j+1)}(t)+e^{t}f^{(j+1)}(0)+\int_0^te^{t-u}f^{(j+2)}(u)du\\ &=\displaystyle -\sum_{j=0}^{k+1}f^{(j)}(t)+e^t\sum_{j=0}^{k+1}f^{(j)}(0)+\int_0^te^{t-u}f^{(j+2)}(u)du, \end{aligned}

which means that this equation holds for all $j\in\mathbb{N}$. On the other hand, $f^{(n+1)}$ is identical to $0$, since $n$ is the degree of $f$. So, the integral will vanish after $n$ such steps, and the assertion follows. q.e.d.

Note that we have vitally used the properties of the number $e$ here, to be precise we used that the function $e^x$ is it’s own derivative. This is the one moment in the proof that hinges specifically on the number $e$ and this is why we will have proved that $e$ is transcendental in the end.

We wish to prove that $e$ is not algebraic. So, let’s assume otherwise: we assume that there is some $g=\sum_{k=0}^rb_kx^k\in\mathbb{Z}[x]$ such that $g(e)=0$. Without loss of generality, we can take $b_0\neq0$ (or else, take out $x$ and use the new polynomial instead of $g$).

Consider $J = \sum_{k=0}^rb_kI(k)$, which equals (by the Integral-Lemma):

\begin{aligned}\displaystyle J = \sum_{k=0}^rb_k e^k\sum_{j=0}^nf^{(j)}(0)-\sum_{k=0}^rb_k\sum_{j=0}^nf^{(j)}(k) &= g(e)\sum_{j=0}^nf^{(j)}(0)-\sum_{k=0}^rb_k\sum_{j=0}^nf^{(j)}(k)\\ &=\displaystyle -\sum_{k=0}^rb_k\sum_{j=0}^nf^{(j)}(k), \end{aligned}

where we used the assumption that $g(e)=0$. Note, that the polynomial $f$ itself has not mattered yet. Let $p$ a sufficiently large prime number with $p>r$ and $p>\left|b_0\right|$ and let us specify $f$ as follows:

$\displaystyle f(x) = x^{p-1}(x-1)^p(x-2)^p\cdots (x-r)^p = x^{p-1}\prod_{k=1}^r(x-k)^p.$

The degree of $f$ is then $n=(r+1)p-1$. We will show that $J$ with this choice of $f$ is divisible by $(p-1)!$, but not divisible by $p$. Each term in $J$ is divisible by $p!$, except for $f^{(p-1)}(0)$.

Let us first deal with the first summand of $J$. The following is true:

The First $p!$-Lemma: $f^{(j)}(0)$ is divisible by $p!$ for any $j\geq p$, and not divisible by $p$ for $j=p-1$.

Proof: By the product rule and the binomial theorem, we have

\begin{aligned} f^{(j)}(0) &=\displaystyle \left. \frac{d^j}{dx^j} x^{p-1}\prod_{k=1}^r(x-k)^p\right|_{x=0} \\ &= \sum_{m=0}^j \binom{j}{m} \left.\frac{d^m}{dx^m}x^{p-1}\right|_{x=0} \left.\frac{d^{j-m}}{dx^{j-m}}\prod_{k=1}^r(x-k)^p\right|_{x=0}. \end{aligned}

Now,

$\displaystyle \frac{d^m}{dx^m}x^{p-1} =\begin{cases} \frac{(p-1)!}{(p-1-m)!} x^{p-1-m},& m\leq p-1,\\ 0&m>p-1,\end{cases}$

and hence

$\displaystyle \left.\frac{d^m}{dx^m}x^{p-1}\right|_{x=0} = \begin{cases}(p-1)!,& m=p-1,\\ 0 &m\neq p-1.\end{cases}$

Therefore, only the case $m=p-1$ is relevant (which immediately implies $j\geq p-1$ to get a non-trivial case), and we still have to take care of the term

$\displaystyle \left.\frac{d^{j-p+1}}{dx^{j-p+1}}\prod_{k=1}^r(x-k)^p\right|_{x=0}.$

The case $j=p-1$ is particularly easy, we arrive at

$\displaystyle \prod_{k=1}^r(-k)^p = (-1)^{rp}(r!)^p.$

In particular, this expression is not divisible by $p$, since we chose a prime $p>r$ from the start.

For the cases $j>p-1$, we use the product rule again, and the multinomial theorem for the first time:

\begin{aligned} \displaystyle \left.\frac{d^{j-p+1}}{dx^{j-p+1}}\prod_{k=1}^r(x-k)^p\right|_{x=0} &= \sum_{m_1+m_2+\cdots+m_r=j-p+1}\binom{j-p+1}{m_1,m_2,\cdots,m_r}\prod_{k=1}^r\frac{p!}{(p-m_k)!}(-k)^{p-m_k}\\ &= \sum_{m_1+m_2+\cdots+m_r=j-p+1}\frac{(j-p+1)!}{m_1!m_2!\cdots m_r!}\prod_{k=1}^r\frac{p!}{(p-m_k)!}(-k)^{p-m_k}\\ &= (j-p+1)!\sum_{m_1+m_2+\cdots+m_r=j-p+1}\prod_{k=1}^r\binom{p}{m_k}(-k)^{p-m_k}\\ &= (j-p+1)!\cdot C_1, \end{aligned}

where $C_1$ is an integer of which the exact value is not important (note: $C_1$ may vanish, for instance if $j$ is sufficiently large). This implies

\begin{aligned}\displaystyle f^{(j)}(0) &= \binom{j}{p-1}(p-1)!(j-p+1)!\cdot C_1 \\&=\displaystyle C_1 \frac{j!}{(p-1)!(j-p+1)!}(p-1)!(j-p+1)! \\&=\displaystyle C_1 j!.\end{aligned}

Altogether, we have shown

$\displaystyle f^{(j)}(0) = \begin{cases}C_1j! & j>p-1,\\(p-1)!(-1)^{rp}(r!)^p, & j=p-1,\\ 0&j

That’s our claim. q.e.d.

The second summand of $J$ works similarly, but the details are a little more cumbersome.

The second $p!$-Lemma: $f^{(j)}(k)$ is divisible by $p!$ for any $j\geq p$ and any $k>0$.

Proof: By the product rule and the binomial theorem, we have

\begin{aligned} f^{(j)}(k) &=\displaystyle \left. \frac{d^j}{dx^j} (x-k)^p\frac1x\prod_{t=0, t\neq k}^r(x-t)^p\right|_{x=k} \\ &= \sum_{m=0}^j \binom{j}{m} \left.\frac{d^m}{dx^m}(x-k)^p\right|_{x=k} \left.\frac{d^{j-m}}{dx^{j-m}}\frac1x\prod_{t=0, t\neq k}^r(x-t)^p\right|_{x=k}. \end{aligned}

Of course,

$\displaystyle \frac{d^m}{dx^m}(x-k)^p = \begin{cases}\displaystyle \frac{p!}{(p-m)!} (x-k)^{p-m}, &m\leq p,\\ 0&m>p-1,\end{cases}$

and so

$\displaystyle \left.\frac{d^m}{dx^m}(x-k)^p\right|_{x=k} = \begin{cases}p!, & m=p,\\ 0 & m\neq p.\end{cases}$

We now have left to take care of the term

$\displaystyle \left.\frac{d^{j-p}}{dx^{j-p}}\frac1x\prod_{t=0, t\neq k}^r(x-t)^p\right|_{x=k}.$

For $j=p$, we find that this is equal to

$\displaystyle \frac1k\prod_{t=0, t\neq k}^r(k-t)^p = k^{p-1}\prod_{t=1,t\neq k}^r(k-t)^p = C_2,$

where $C_2$ is an integer (note that $p$ is prime, hence $p\geq2$ and hence there will be no fraction in this term). In particular, $f^{(p)}(k) = C_2p!$ is divisible by $p!$.

For $j>p$, we can apply the multinomial theorem once again to find

\begin{aligned} \displaystyle \frac{d^{j-p}}{dx^{j-p}}\frac1x\prod_{t=0, t\neq k}^r(x-t)^p &= \sum_{m_0=0}^{j-p}\binom{j-p}{m_0}\left.\frac{d^{m_0}}{dx^{m_0}}x^{p-1}\right|_{x=k}\left.\frac{d^{j-p-m_0}}{dx^{j-p-m_0}}\prod_{t=1, t\neq k}^r(x-t)^{p}\right|_{x=k}\\ &= \sum_{m_0=0}^{j-p}\binom{j-p}{m_0}\frac{(p-1)!}{(p-1-m_0)!}k^{p-1-m_0}\times\\ &\displaystyle\hphantom{=}\times\sum_{m_1+\cdots+m_r=j-p-m_0, m_k=0}\binom{j-p-m_0}{m_1,\cdots,m_r}\prod_{t=1, t\neq k}^{r}\frac{p!}{(p-m_t)!}(k-t)^{p-m_t}\\ &=\sum_{m_0=0}^{j-p}\binom{j-p}{m_0}\frac{(p-1)!}{(p-1-m_0)!}k^{p-1-m_0}\times\\ &\displaystyle\hphantom{=}\times\sum_{m_1+\cdots+m_r=j-p-m_0, m_k=0}\frac{(j-p-m_0)!}{m_1!\cdots m_r!}\prod_{t=1, t\neq k}^{r}\frac{p!}{(p-m_t)!}(k-t)^{p-m_t}\\ &=\sum_{m_0=0}^{j-p}\frac{(j-p)!}{m_0!(j-p-m_0)!}\frac{(p-1)!}{(p-1-m_0)!}k^{p-1-m_0}(j-p-m_0)!\times\\ &\displaystyle\hphantom{=}\times\sum_{m_1+\cdots+m_r=j-p-m_0, m_k=0}\prod_{t=1, t\neq k}^{r}\binom{p}{m_t}(k-t)^{p-m_t}\\ &=(j-p)!\sum_{m_0=0}^{j-p}\binom{p-1}{m_0}k^{p-1-m_0}\times\\ &\displaystyle\hphantom{=}\times\sum_{m_1+\cdots+m_r=j-p-m_0, m_k=0}\prod_{t=1, t\neq k}^{r}\binom{p}{m_t}(k-t)^{p-m_t}. \end{aligned}

This means, that we have proved

\begin{aligned} f^{(j)}(k) &= \binom{j}{p}p!\left.\frac{d^{j-p}}{dx^{j-p}}\frac1x\prod_{t=0, t\neq k}^r(x-t)^p\right|_{x=k}\\ &= \frac{j!}{p!(j-p)!}p!(j-p)!\cdot C_3\\ &= C_3 j!, \end{aligned}

where $C_3$ is some integer which is of no importance in itself (note: $C_3$ may vanish, if $j$ is sufficiently large for instance – in the pathological cases, the binomial and multinomial coefficients vanish by definition). This is what we claimed: $f^{(j)}$ is divisible by $p!$ for any $j> p$ and hence for any $j\geq p$. q.e.d.

Where do we stand? We consider the expression

$\displaystyle J = -\sum_{k=0}^r\sum_{j=0}^nb_kf^{(j)}(k),$

where $f$ is a certain polynomial that has multiple roots in $k=0,\ldots,r$. As in the proof of both $p!$-Lemmas, we find that $f^{(j)}(0) = 0$ for any $j< p-1$ and $f^{(j)}(k) = 0$ for any $j and $k=1,\ldots,r$. Hence

$\displaystyle J = -\sum_{j=p-1}^nb_0f^{(j)}(0) + \sum_{k=1}^r\sum_{j=p}^nb_kf^{(j)}(k)$.

The second $p!$-Lemma showed, that the second summand is divisible by $p!$. The first $p!$-Lemma showed that $f^{(j)}(0)$ is only divisible by $(p-1)!$, not by $p$. Now, since we chose $p>\left|{b_0}\right|$ in the start, the whole first summand of $J$ is not divisible by $p!$. On top of all that, $J$ is an integer, since the $b_k$ are by assumption and the other terms are by construction and by the Lemmas. Altogether, we conclude $J$ is a multiple of $(p-1)!$, but not divisible by $p$, and hence

$\left| J\right| \geq (p-1)!$.

Now, we give an upper bound for $\overline f(k)$ for any $k=0,\ldots,r$:

$\displaystyle \overline f(k) = k^{p-1}\prod_{t=1}^rt^p\leq (2r)^{p-1}\prod_{t=1}^r(2r)^p = (2r)^{p-1+rp} =\frac12 ((2r)^{r+1})^p.$

From the immediate bound we gave at the very beginning,

\begin{aligned} \displaystyle \left|J\right| &\leq\sum_{j=0}^r\left|b_j\right|\left|I(j)\right| \\ &\displaystyle \leq \sum_{j=0}^r \left| b_j\right| \left|t\right|e^{\left|t\right|}\overline f(\left|t\right|) \\ &\displaystyle \leq \sum_{j=0}^r\left| b_j\right| \left|t\right|e^{\left|t\right|}\frac12 ((2r)^{r+1})^p \\ &\displaystyle \leq ((2r)^{r+1})^p \frac12 \left|t\right|e^{\left|t\right|}\sum_{j=0}^r\left| b_j\right|\\ &\displaystyle \leq C_1 s^{p}. \end{aligned}

with the unimportant constants $C_1= \frac12\left|t\right|e^{\left|t\right|}\sum_{j=0}^r\left|b_j\right|$ and $s=(2r)^{r+1}$ that do not depend on $p$.

So, on the one hand, $\left|J\right|$ is large, (larger than $(p-1)!$), on the other hand it’s small (smaller than $C_1 x^p$). This is impossible, because $(p-1)!$ eventually gets larger than $x^p$ — this can be seen by considering the exponential sequence $\frac{x^n}{n!}$ that is summable and hence converges to $0$. In particular, $\frac{x^p}{p!}\leq 1$ eventually, which means $x^p\leq p!$. There’s the contradiction: $J$ cannot satisfy both inequalities that we have shown. The contradiction has its roots in our assumption — $e$ must be transcendental. q.e.d.

We shall now prove

Theorem: $\pi$ is transcendental.

In this proof we will use many of the ideas from the proof about $e$, and therefore we will not be as detailed as above. Besides, the proof requires some results from algebra courses. We will briefly review those here.

The number $z\in\mathbb{C}$ is called an algebraic integer, if its minimal polynomial has coefficients in $\mathbb{Z}$. The minimal polynomial is the (unique) polynomial of lowest degree which has $z$ as a root and whose leading coefficient is equal to $1$.

The Integer Lemma:

1. Any algebraic integer $z\in\mathbb{Q}$ is already in $\mathbb{Z}$.
2. If $z\in\mathbb{C}$ is algebraic then there is some $b\in\mathbb{N}$ such that $bz$ is an algebraic integer.

Proof: 1. Any $z\in\mathbb{Q}$ is algebraic, because its minimal polynomial is $x-z\in\mathbb{Q}[x]$. By assumption, $z$ is an algebraic integer, hence its minimal polynomial is in $\mathbb{Z}[x]$. Hence, $z\in\mathbb{Z}$, as claimed.

2. Let us designate the minimal polynomial of $z$ by $g\in\mathbb{Q}$. Let us put

$\displaystyle g(x) = \sum_{k=0}^nq_kx^k,\qquad q_k\in\mathbb{Q}$.

Any such coefficient $q_k$ can be written as $\frac{s_k}{t_k}$ for appropriate $s_k\in\mathbb{Z}$ and $t_k\in\mathbb{N}$. Let $q:=lcm(t_1,\ldots,t_n)$ so we can get

$\displaystyle qg(x)=\sum_{k=0}^nc_kx^k,\qquad c_k\in\mathbb{Z}$.

This polynomial $qg$ is not necessarily a minimal polynomial, since its leading coefficient is no longer equal to $1$, in general. But, on the other hand, we find

latex \begin{aligned} 0 = g(z) = c_n^{n-1}qg(z) &=\displaystyle\sum_{k=0}^n c_n^{n-1}c_kz^k\\ &= \displaystyle \sum_{k=0}^{n-1} c_n^{n-1-k} c_k c_n^kz^k + c_n^{n-1}c_nz^n\\&= \displaystyle \sum_{k=0}^{n-1} c_n^{n-1-k} c_k (c_nz)^k + (c_nz)^n\\&= \displaystyle \sum_{k=0}^n d_k (c_nz)^k\\&=: \tilde g(c_nz).\end{aligned}

Thus, we have found a polynomial $\tilde g\in\mathbb{Z}[x]$ which has $c_nz$ as a root and which has leading coefficient $1$. This means $c_nz$ is an algebraic integer. A closer inspection shows that $c_n$ is equal to $q$ which, in turn, was the lcm of several positive integers, $c_n$ itself is positive. Hence we can choose $b:=c_n$ in the claim and we are done. q.e.d.

The Lemma on Algebraic Numbers (aka the LAN): If $\alpha$ and $\beta$ are algebraic, then so are $\alpha\pm\beta$, $\alpha\beta$ and $\frac\alpha\beta$.

Proof: This follows from a theorem in field theory stating that in a field extension $K\subset K(\alpha)$, $\alpha$ is algebraic if and only if its degree is finite. This can easily be seen as follows:

In a field extension $K\subset K(\alpha)$ of degree $n<\infty$, the elements $1,\alpha,\ldots,\alpha^n$ are linear dependent, so we find $\sum_{i=0}^na_i\alpha^i=0$. This is a polynomial which has $\alpha$ as its root, so $\alpha$ is algebraic. On the other hand, if $\alpha$ is algebraic and $n$ is the degree of its minimal polynomial, any $\beta\in K(\alpha)$ has a representation as $\beta=\sum_{i=0}^{n-1}b_i\alpha^i$ – keep in mind that $K(\alpha)$ is a vector space over $K$ and the set $\{1,\alpha,\ldots,\alpha^n\}$ must be linear dependent. In particular, the degree of the extension is finite.

Now the usual theorem on degrees of extensions tells us that any finite field extension is algebraic, because if $K\subset L$ is a finite extension of degree $n$, and $\alpha\in L$:

$\displaystyle [L:K(\alpha)] [K(\alpha):K] = [L:K]<\infty$,

and so $K(\alpha)\subset K$ is finite, hence $\alpha$ is algebraic. Note that the converse of this is false ($\mathbb{Q}\subset\overline{\mathbb{Q}}$).

Now consider the algebraic numbers $\alpha$ and $\beta$ and the field extension $K\subset K(\alpha, \beta)$. By the arguments above, the extension is finite (it’s a tower of two finite extensions) and hence every element in $K(\alpha,\beta)$ is algebraic. The lemma is proved. q.e.d.

Now we can turn to the proof that $\pi$ is transcendental. But we won’t prove that directly, instead we show that $i\pi$ is transcendental. Because of the LAN, if $i\pi$ is algebraic, then so is $\pi$.

Let us assume that $\theta:=i\pi$ is algebraic with minimal polynomial $g\in\mathbb{Q}[x]$ and degree $r$. We will designate with $b$ the leading coefficient of $\tilde g\in\mathbb{Z}[x]$ that arises from getting rid of the denominators in $g$. Let $\theta_1=\theta$, $\theta_2,\ldots, \theta_r$ be the roots of $g$. Then, by the Integer Lemma, $b\theta_j$ is an algebraic integer. It is immediate that

$\displaystyle \prod_{k=1}^r(1+e^{\theta_k}) = 0,$

because of $e^{\theta_1}=e^{i\pi} = -1$. By multiplication of this product, we get $2^r$ terms that look like $e^{\phi}$ where $\phi = \sum_{k=1}^r\epsilon_k\theta_k$ and $\epsilon_k\in\{0,1\}$ (this comes from the fact that you have to choose either $e^{\theta_k}$ or $1$ from each factor, and each of the $2^r$ possible choices will be made once).

Let us denote the non-zero summands by $\phi_k$ and the number of non-zero summands by $n$. So, there are $2^r-n$ summands that show up as $e^0=1$. Altogether,

$\displaystyle 2^r-n+\sum_{k=1}^ne^{\phi_k} = 0\qquad (\clubsuit)$.

Now we re-use some ideas from the proof that $e$ is transcendental. Let $p$ be a large prime number and let

$\displaystyle f(x) = b^{np}x^{p-1}\prod_{k=1}^n(x-\phi_k)^p.$

Here, consider the expression

$\displaystyle \prod_{k=1}^{2^r}(x-\phi_k) = x^{2^r-n}\prod_{k=1}^n(x-\phi_k)$

and note that it is symmetric in the $\theta_k$: when you re-number the $\theta_k$, you will also re-number the $\phi_k$, but the polynomial given here will not change. The Fundamental Theorem of Elementary Symmetric Polynomials says that this symmetric polynomial can be written as a polynomial in the elementary symmetric polynomials with coefficients in the underlying field $\mathbb{Q}$. In particular, the polynomial

$\displaystyle \prod_{k=1}^n(x-\phi_k)$

has coefficients in $\mathbb{Q}$, the remaining factor $x^{2^r-n}$ does not affect any of the coefficients. Besides, the coefficients are algebraic by assumption and by the LAN (the $\theta_k$ are algebraic). By the Integer Lemma (closely inspect the proof for the possible choice of the number $b$ in that Lemma!) and by the definition of $b$ as a leading coefficient here, the numbers $b\phi_k$ are algebraic integers. The second part of the Integer Lemma now says that the polynomial $f$ has coefficients in $\mathbb{Z}$ – each of the factors of the symmetric polynomial $\prod_{k=1}^n(x-\phi_k)$ is multiplied by $b$ to make the coefficients in $\mathbb{Q}$ algebraic integers.

Now, remember the definition of $I$ and its representation from the Integral Lemma:

$\displaystyle I(t) = e^t\sum_{j=0}^mf^{(j)}(0)-\sum_{j=0}^mf^{(j)}(t)$,

where $m=(n+1)p-1$ is the degree of $f$ as defined above. Then, we set

\begin{aligned} \displaystyle J := \sum_{k=1}^nI(\phi_k) &\overset{\hphantom{(\clubsuit)}}{=} \sum_{k=1}^n e^{\phi_k}\sum_{j=0}^mf^{(j)}(0)-\sum_{k=1}^n\sum_{j=0}^mf^{(j)}(\phi_k)\\ &\displaystyle \overset{(\clubsuit)}{=} -(2^r-n)\sum_{j=0}^mf^{(j)}(0)-\sum_{j=0}^m\sum_{k=1}^nf^{(j)}(\phi_k). \end{aligned}

Similar reasoning now shows that the term

$\displaystyle \sum_{k=1}^nf^{(j)}(\phi_k)$

is a symmetric polynomial: when you re-number the $\theta_k$ and hence the $\phi_k$, $f$ will not change, and by the product rule, neither will the derivatives $f^{(j)}$. By the Fundamental Theorem of Elementary Symmetric Polynomials, this sum takes a value in the underlying field $\mathbb{Q}$. Again, the Integer Lemma implies that the sum is even in $\mathbb{Z}$ (the helpful factor $b$ is of course present in the relevant derivatives as well).

Now, remember the first $p!$-Lemma to see that $f^{(j)}(\phi_k)=0$ for any $j and any $k$. This hinges on the fact that none of the factors $(x-\phi_k)$ can have vanished in the $j$-th derivative, and then setting $x=\phi_k$ gives a zero of this polynomial. By the same argument, $f^{(j)}(0)=0$ for $j and by the second $p!$-Lemma, $f^{(j)}(0)$ is divisible by $p!$ for $j\geq p$. Finally, one can show by explicit calculation

$\displaystyle f^{(p-1)}(0) = b^{np}(-1)^{np}(p-1)!\prod_{k=1}^n\phi_k^p.$

Since this expression is symmetric, it is in $\mathbb{Q}$, and from the arguments we already used, $f^{(p-1)}(0)\in\mathbb{Z}$. The symmetry arguments also show that $J$ is an integer in $\mathbb{Z}$. If we choose the prime $p$ sufficiently large, $f^{(p-1)}$ is not divisible by $p$. This is a condition depending on $b$ and on the absolute value of the $\phi_k$, hence on the degree $r$ of $g$. In particular, the sum $J$ is divisible by $(p-1)!$.

$\left|J\right|\geq(p-1)!$.

Finally, the bound from the Integral Lemma tells us that

$\displaystyle \left|J\right| \leq\sum_{k=1}^n\left|I(\phi_k)\right| \leq\sum_{k=1}^n\left|\phi_k\right|e^{\left|\phi_k\right|}\overline{f}(\left|\phi_k\right|)\leq C_1C_2^p,$

with two constants $C_1, C_2$ that do not depend on $p$. The exponent $p$ comes from the definition of the polynomial $f$.

Similar to the proof of our first theorem, we have a contradiction: $J$ is small on the one hand and large on the other hand. This shows that $\pi$ cannot be algebraic. q.e.d.