Introduction
The Euler-Napier-Bernoulli constant e =2.71828\ldots is not just irrational, it is transcendental, as first proved by Charles Hermite in 1873. Inspired by the work of Mathologer (Burkard Polster with Marty Ross), here I offer an elementary proof of e‘s transcendence. As warmup, I first present a well-known proof of its irrationality while hinting at the proof of its transcendence.
e is irrational
The infinite series expansion of the exponential function
e^x = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \cdots + \frac{x^m}{m!} + \cdots
implies the Euler-Napier-Bernoulli constant
e = 1 + 1 + \frac{1}{2!} + \frac{1}{3!} + \cdots + \frac{1}{m!} + \cdots .
By way of contradiction, assume e can be written as a root of the linear polynomial
be-a = 0
for positive integers a,b \in \mathbb{Z}^+ or, equivalently, that e is the rational number
e = \frac{a}{b}.
Use b as a cutoff to separate e‘s infinite series expansion into a \color{blue}\text{large body} large body and a \color{red}\text{small tail}
\frac{a}{b} = e = \color{blue} 1 + 1 + \frac{1}{2!} + \frac{1}{3!} + \cdots + \frac{1}{b!} \color{red}+ \frac{\delta}{b!}\color{black},
where
\begin{aligned}0 < \delta &= \frac{b!}{(b+1)!} + \frac{b!}{(b+2)!} + \frac{b!}{(b+3)!} + \cdots \\&= \frac{1}{b+1} + \frac{1}{(b+1)(b+2)} + \frac{1}{(b+1)(b+2)(b+3)} + \cdots \\&< \frac{1}{b+1} + \frac{1}{(b+1)^2} + \frac{1}{(b+1)^3} + \cdots \\&= \frac{1/(b+1)}{1-1/(b+1)} = \frac{1}{b} \le 1\end{aligned}
using the sum formula for an infinite geometric series. But
\frac{a}{b} = e = \frac{b!\left(1 + 1 + \frac{1}{2!} + \frac{1}{3!} + \cdots + \frac{1}{b!}\right) + \delta}{b!} \color{red} = \frac{N^\prime + \delta}{N}
and so
a\color{red}N\color{black}-b\color{red}N^\prime\color{black}-b\color{red}\delta\color{black}=0
and
a(b-1)!-b!\left(1 + 1 + \frac{1}{2!} + \frac{1}{3!} + \cdots + \frac{1}{b!}\right)-\delta = 0,
which is impossible, as all terms are integers accept 0 < \delta <1. Hence e is not the root of a linear polynomial with integer coefficients and, equivalently, is an irrational number.
e is transcendental
First consider the polynomial
f(x) = \color{red}x^{p-1}\color{black}(x-1)^p(x-2)^p \cdots (x-n)^p,
where n > 1 and p is a large prime. As |x-m|\le n on the interval x\in [0,n], it is bounded by
|f| \le \color{red}n^{p-1}\color{black}n^p n^p \cdots n^p = n^{np+p-1},
and its expansion begins
\begin{aligned}f(x) &= (-1)^p(-2)^p\cdots (-n)^p \color{red}x^{p-1}\color{black} + \mathcal{O}(x^{p})\\&= (-1)^{np}(n!)^p \color{red}x^{p-1}\color{black} + \mathcal{O}(x^{p}).\end{aligned}
Use the factorial integral
\int_0^\infty \hspace{-0.8em}\text{d}x \,e^{-x} x^m = m!
to expand powers of e like
e^m = \frac{ \int_0^\infty \hspace{-0.4em}\text{d}x \,e^{m-x} f(x)}{ \int_0^\infty \hspace{-0.4em}\text{d}x \,e^{-x} f(x)} \color{red}=\ \frac{N_m + \delta_m}{N}\color{black},
where
\begin{aligned}N &=\frac{1}{(p-1)!} \int_0^\infty \hspace{-0.8em}\text{d}x \,e^{-x} f(x)\\&= (-1)^{np}(n!)^p + M p,\end{aligned}
where M \in \mathbb{Z} and so N \in \mathbb{Z}, and if p > n then N\neq 0, as p divides the second term but not the first,
\begin{aligned}N_m &=\frac{1}{(p-1)!} \int_m^\infty \hspace{-0.8em}\text{d}x \,e^{m-x} f(x)\\&= \frac{1}{(p-1)!} \int_0^\infty \hspace{-0.8em}\text{d}y \,e^{-y} f(y+m) = M^\prime p,\end{aligned}
where M^\prime \in \mathbb{Z}, as f(y+m) has a factor of y^p enabling the integral to absorb the denominator, and
\begin{aligned}\delta_m &=\frac{1}{(p-1)!}\int_0^m \hspace{-0.8em}\text{d}x \,e^{m-x}f(x)\\&\le \frac{1}{(p-1)!} \color{red}e^m\int_0^\infty \hspace{-0.8em}\text{d}x \,e^{-x}\color{blue} n^{np+p-1}\color{black}\\&=\frac{1}{(p-1)!} \color{red}e^m\color{blue}\frac{n^{(n+1)p}}{n}\color{black}\\&\le \frac{1}{(p-1)!} \color{red}e^n\color{blue} n^{(n+1)p}\color{black} = \frac{c\,d^p}{(p-1)!} \rightarrow 0,\end{aligned}
as p\rightarrow \infty, where c = e^n and d = n^{n+1}>1.
Now, by way of contradiction, assume e can be written as a root of the polynomial
a_n e^n+a_{n-1}e^{n-1}+\cdots+a_0 = 0
of lowest degree with a_m\in\mathbb{Z}, for m=1,\ldots,n, and a_0\neq 0. Replace powers of e to get
a_n \left(\frac{N_n+\delta_n}{N}\right)+a_{n-1}\left(\frac{N_{n-1}+\delta_{n-1}}{N}\right)+\cdots+a_0 = 0
and so
a_0 \color{red}N\color{black}+(a_1 \color{red}N_1\color{black} + \cdots + a_n \color{red}N_n\color{black}) + (a_1 \color{red}\delta_1\color{black} + \cdots + a_n\color{red}\delta_n\color{black}) = 0,
which is impossible, as the N and N_m terms are integers but the \delta_m terms are arbitrarily small. (For large enough p >|a_0| and p >n, p does not divide a_0 N but does divide N_m, and the integer parts cannot vanish.) Hence e is not the root of a polynomial with integer coefficients and, therefore, is a transcendental number.
Thanks, Mark! I enjoy reading your posts as well.