Famous Theorems of Mathematics/e is transcendental

The mathematical constant $$e=\sum_{n\,=\,0}^\infty\frac{1}{n!}=2.718281\ldots$$ is a transcendental number.

In other words, it is not a root of any polynomial with integer coefficients.

Proof
Let us assume that $$e$$ is algebraic, so there exists a polynomial such that
 * $$P(e)=a_0+a_1e+a_2e^2+\cdots+a_ne^n=0$$

for $$a_0,\ldots,a_n\in\Z$$ and $$a_0\ne0$$.

Part 1
Let $$f(x)$$ be a polynomial of degree $$d$$. Let us define $$F(x)=\sum_{k\,=\,0}^df^{(k)}\!(x)$$. Taking its derivative yields:
 * $$F'\!(x)=\sum_{k\,=\,0}^df^{(k+1)}\!(x)=\sum_{k\,=\,1}^df^{(k)}\!(x)=F(x)-f(x)$$

Let us define $$G(x)=e^{-x}F(x)$$. Taking its derivative yields:
 * $$G'\!(x)=e^{-x}F'\!(x)-e^{-x}F(x)=e^{-x}\bigl[F'\!(x)-F(x)\bigr]=-e^{-x}f(x)$$

Since $$G(x)$$ is differentiable, we shall apply the mean Value Theorem on the interval $$[0,m]$$ for $$m\in\N$$. So there exists a $$x_m\in(0,m)$$ such that
 * $$G'\!(x_m)=\frac{G(m)-G(0)}{m-0}=\frac{e^{-m}F(m)-e^{-0}F(0)}{m}=-e^{-x_m}f(x_m)$$

Now let:
 * $$\begin{align}&F(m)-e^mF(0)=-m\,e^{m-x_m}f(x_m)=A_m\\[6pt]&a_mF(m)-a_me^mF(0)=a_mA_m\end{align}$$

Summing all the terms yields
 * $$\begin{align}\sum_{m\,=\,1}^na_mF(m)-\sum_{m\,=\,1}^na_me^mF(0)=\sum_{m\,=\,1}^na_mA_m\\[6pt]\sum_{m\,=\,1}^na_mF(m)-F(0)\sum_{m\,=\,1}^na_me^m=\sum_{m\,=\,1}^na_mA_m\\[6pt]\sum_{m\,=\,1}^na_mF(m)-F(0)(-a_0)=\sum_{m\,=\,1}^na_mA_m\\[6pt]\sum_{m\,=\,0}^na_mF(m)=\sum_{m\,=\,1}^na_mA_m\end{align}$$

Part 2
Let $$f(x)$$ be a polynomial with a root $$x_0$$ of multiplicity $$d$$. We will show that for all $$0\le k\le d-1$$ we get $$f^{(k)}\!(x_0)=0$$.

Let us write $$f(x)=(x-x_0)^dQ_0(x)$$, such that $$Q_0(x)$$ is a polynomial with $$Q_0(x_0)\ne0$$.
 * $$\begin{align}f^{(1)}\!(x)&=d(x-x_0)^{d-1}Q_0(x)+(x-x_0)^dQ_0'(x)\\[5pt]&=(x-x_0)^{d-1}\Big[d\,Q_0(x)+(x-x_0)Q_0'(x)\Big]\\[5pt]&=(x-x_0)^{d-1}Q_1(x)\\[5pt]Q_1(x_0)&=d\,Q_0(x_0)\ne0\\\\[5pt]f^{(2)}\!(x)&=(d-1)(x-x_0)^{d-2}Q_1(x)+(x-x_0)^{d-1}Q_1'(x)\\[5pt]&=(x-x_0)^{d-2}\Big[(d-1)Q_1(x)+(x-x_0)Q_1'(x)\Big]\\[5pt]&=(x-x_0)^{d-2}Q_2(x)\\[5pt]Q_2(x_0)&=(d-1)Q_1(x_0)\ne0\\[5pt]&\,\,\,\vdots\\[5pt]f^{(d-1)}\!(x)&=2(x-x_0)Q_{d-2}(x)+(x-x_0)^2Q_{d-2}'(x)\\[5pt]&=(x-x_0)\Big[2Q_{d-2}(x)+(x-x_0)Q_{d-2}'(x)\Big]\\[5pt]&=(x-x_0)Q_{d-1}(x)\\[5pt]Q_{d-1}(x_0)&=2Q_{d-2}(x_0)\ne0\end{align}$$

with $$Q_0'(x),Q_1'(x),\ldots,Q_{d-1}'(x)$$ all polynomials.

Part 3
Let us now define a polynomial
 * $$\begin{align}f(x)&=\frac{1}{(p-1)!}\,x^{p-1}\bigl[(1-x)(2-x)\cdots(n-x)\bigr]^p\\[5pt]&=\frac{(n!)^p}{(p-1)!}x^{p-1}+\!\!\sum_{m\,=\,p}^{(n+1)p-1}\!\!\!\frac{b_m}{(p-1)!}x^m\quad:b_m\in\Z\end{align}$$

for prime $$p$$ such that $$p>a_0$$ and $$p>n$$. We get:
 * $$f^{(k)}\!(x)=\!\!\sum_{m\,=\,k}^{(n+1)p-1}\!\!\!\frac{k!}{(p-1)!}\binom{m}{k}b_mx^{m-k}\quad:p\le k\le(n+1)p-1$$

hence for all $$k\ge p$$, the function $$f^{(k)}\!(x)$$ is a polynomial with integer coefficients all divisible by $$p$$.

By part 2, for all $$1\le m\le n$$ we get:
 * $$F(m)=\!\!\sum_{k\,=\,0}^{(n+1)p-1}\!\!\!\!f^{(k)}\!(m)=\!\!\sum_{k\,=\,p}^{(n+1)p-1}\!\!\!\!f^{(k)}\!(m)$$

Therefore $$F(m)$$ is also an integer divisible by $$p$$.

On the other hand, for $$m=0$$ we get:
 * $$F(0)=\!\!\sum_{k\,=\,0}^{(n+1)p-1}\!\!\!\!f^{(k)}\!(0)=\!\!\sum_{k\,=\,p-1}^{(n+1)p-1}\!\!\!\!f^{(k)}\!(0)$$

but $$f^{(p-1)}\!(0)=(n!)^p$$, and $$n,a_0$$ are not divisible by $$p$$. Therefore $$a_0F(0)$$ is not divisible by $$p$$.

In other words, the sum $$\sum_{m\,=\,0}^na_mF(m)$$ is an integer not divisible by $$p$$, and particularly non-zero.

Part 4
By part 1, for all $$1\le m\le n$$ we have $$0<x_m<m\le n$$. Therefore,
 * $$\begin{align}A_m&=-m\,e^{m-x_m}f(c_m)=-\frac{m\,e^{m-x_m}}{(p-1)!}\,(x_m)^{p-1}\bigl[(1-x_m)(2-x_m)\cdots(n-x_m)\bigr]^p\\[5pt]|A_m|&=\frac{m\,e^{m-x_m}}{(p-1)!}\,(x_m)^{p-1}\Big|(1-x_m)(2-x_m)\cdots(n-x_m)\Big|^p\\[5pt]&\le\frac{ne^n}{(p-1)!}\,n^{p-1}(1\cdot2\cdots n)^p=e^n\frac{(n\cdot n!)^p}{(p-1)!}\end{align}$$

By the triangle Inequality we get:
 * $$0<\left|\sum_{m\,=\,0}^na_mF(m)\right|=\left|\sum_{m\,=\,1}^na_mA_m\right|=\sum_{m\,=\,1}^n|a_mA_m|\le\left(\sum_{m\,=\,1}^n|a_m|\right)e^n\frac{(n\cdot n!)^p}{(p-1)!}$$

But $$\lim_{p\to\infty}\frac{(n\cdot n!)^p}{(p-1)!}=0$$, hence for sufficiently large $$p$$ we get $$0<\left|\sum_{m\,=\,0}^na_mF(m)\right|<1$$. A contradiction.

$$\blacksquare$$