e is transcendental

Assume that $e\in\overline{\mathbb{Q}}$, then

$$\begin{eqnarray*} &&\exists a_n(\neq 0),a_{n-1},\cdots,a_1,a_0\in \mathbb{Z},\\ &&a_n e^n+a_{n-1} e^{n-1}+\cdots a_1 e+a_0=0. \end{eqnarray*}$$

Let

$$\begin{eqnarray*} &J&\stackrel{\mathrm{def}}{=}\tsum_{k=0}^n a_k I(k) =\tsum_{k=0}^n a_k \int_0^k e^{k-u}f(u)du, \end{eqnarray*}$$

where

$$\begin{eqnarray*} &f(x)&\stackrel{\mathrm{def}}{=}x^{p-1}(x-1)^p\cdots(x-n)^p,\... ...rline{f}(x)&=x^{p-1}(x+1)^p\cdots(x+n)^p,\\ &m&=\deg f=p-1+np. \end{eqnarray*}$$

Then according to the equality (2) of the lemma 2

$$\begin{eqnarray*} &J&=\tsum_{k=0}^n a_k \left(e^k\tsum_{j=0}^m f^{(j)}(0)-\tsu... ...(k)\right)\\ & &=0-\tsum_{k=0}^n \tsum_{j=0}^m a_k f^{(j)}(k). \end{eqnarray*}$$

Then according to the lemma 3 (the 1st case, see (4)),

$$f^{(j)}(k)\begin{cases} =(p-1)!(-1)^{pn}(n!)^p&\text{(when... ...=0$),}\\ \equiv 0 \pmod{p!} &\text{(otherwise).} \end{cases}$$

So if $p$ is a prime and $p>\max\{\vert a_k\vert,n\}$, then $(p-1)!\mid J$ and $p\nmid J$. Therefore

$$\begin{eqnarray*} &&\vert J\vert\geqq (p-1)!. \end{eqnarray*}$$

On the other hand, according to the inequality (3) of the lemma 2

$$\begin{eqnarray*} &\vert J\vert&\leqq \tsum_{k=0}^n \vert a_k\vert\vert I(k)\ve... ...\max_{0\leqq k\leqq n}\{ \vert a_k\vert k e^k \} (2n)^{(n+1)p}. \end{eqnarray*}$$

The lower and upper bound of $\vert J\vert$ contradict to the lemma 1 (see (1)). Therefore $e$ is transcendental.

pi is transcendental e and pi are transcendental Preparation