Euler's Number is Transcendental

Theorem
Euler's Number $e$ is transcendental.

Proof
Suppose there exist integers $a_0, \ldots, a_n$ with $a_0 \ne 0$ such that:


 * $(1): \quad a_n e^n + a_{n - 1} e^{n - 1} + \cdots + a_0 = 0$

Define $M$, $M_1, \ldots, M_n$ and $\epsilon_1, \ldots, \epsilon_n$ as follows:


 * $\displaystyle M = \int_0^\infty \frac {x^{p - 1} \left[{\left({x - 1}\right) \cdots \left({x - n}\right)}\right]^p e^{- x} } {\left({p - 1}\right)!} \mathrm d x$


 * $\displaystyle M_k = e^k \int_k^\infty \frac {x^{p - 1} \left[{\left({x - 1}\right) \cdots \left({x - n}\right)}\right]^p e^{- x} } {\left({p - 1}\right)!} \mathrm d x$


 * $\displaystyle \epsilon_k = e^k \int_0^k \frac {x^{p - 1} \left[{\left({x - 1}\right) \cdots \left({x - n}\right)}\right]^p e^{- x} } {\left({p - 1}\right)!} \mathrm d x$

where $p$ is a prime number with $p > n$ and $p > \left \vert{a_0}\right \vert$.

The expression $\left[{\left({x - 1}\right) \cdots \left({x - n}\right)}\right]^p$ is a polynomial of degree $n p$ with integer coefficients.

Hence:

where $C_{\alpha}$ are integers and $C_0 = \pm \left({n!}\right)^p$.

For $\alpha = 0$ we have:


 * $C_0 \dfrac {\left({p - 1}\right)!} {\left({p - 1}\right)!} = \pm \left({n!}\right)^p$

Since $p > n$, it follows from Prime iff Coprime to all Smaller Positive Integers and Euclid's Lemma that this term is not divisible by $p$.

For $\alpha \ge 1$ we have:


 * $C_{\alpha} \dfrac {\left({p - 1 + \alpha}\right)!} {\left({p - 1}\right)!} = C_{\alpha} \left({p - 1 + \alpha}\right) \left({p - 2 + \alpha}\right) \cdots p$

which is clearly divisible by $p$.

It follows from Common Divisor Divides Difference that $M$ is an integer not divisible by $p$.

We also have:

The expression $\left[{\left({x + k - 1}\right) \cdots \left({x + k - n}\right)}\right]$ is divisible by $x$.

So $\left({x + k}\right)^{p - 1} \left[{\left({x + k - 1}\right) \cdots \left({x + k - n}\right)}\right]^p$ is a polynomial of degree at least $p$ with integer coefficients.

Hence:

where $D_{\alpha}$ are integers.

Since this sum begins with $\alpha = 1$, each term is divisible by $p$.

Thus each $M_k$ is an integer divisible by $p$.

By the above definitions we have:


 * $e^k = \dfrac {M_k + \epsilon_k} M$

Substituting this into $(1)$ and multiplying by $M$ we obtain:


 * $\left({a_0 M + a_1 M_1 + \cdots + a_n M_n}\right) + \left({a_1 \epsilon_1 + \cdots + a_n \epsilon_n}\right) = 0$

Since $p > \left \vert{a_0}\right \vert$, it follows from Prime iff Coprime to all Smaller Positive Integers that $p$ does not divide $a_0$.

So by Euclid's Lemma, $a_0 M$ is not divisible by $p$.

Since each $M_k$ is not divisible by $p$, it follows from Common Divisor Divides Difference that $a_0 M + a_1 M_1 + \cdots + a_n M_n$ is not divisible by $p$.

Therefore $a_0 M + a_1 M_1 + \cdots + a_n M_n$ is a non-zero integer.

We also have:

Let $A$ be the maximum value of $\left \vert{\left({x - 1}\right) \cdots \left({x - n}\right)}\right \vert$ for $x$ in the interval $\left[{0 \,.\,.\, n}\right]$.

Then:

By Power over Factorial:


 * $\displaystyle \lim_{p \to \infty} \frac {e^n \left({n A}\right)^p} {p!} = 0$

So $\left \vert{\epsilon_k}\right \vert$, and therefore $\left \vert{a_1 \epsilon_1 + \cdots + a_n \epsilon_n}\right \vert$ can be made arbitrarily small by choosing $p$ sufficiently large.

It follows that $\left({a_0 M + a_1 M_1 + \cdots + a_n M_n}\right) + \left({a_1 \epsilon_1 + \cdots + a_n \epsilon_n}\right)$ is non-zero.

This is a contradiction, so $e$ must be transcendental.

Historical Note
The transcendence of $e$ was first proved by Charles Hermite in 1873.

The proof given here is a simplification of Hermite's original proof, given by Hilbert.