Euler's Number is Irrational

Theorem
Euler's number $e$ is irrational.

Proof by Contradiction
Assume that $e$ is rational. Then, there exist coprime integers $m$ and $n$ such that:

$\displaystyle \frac m n = e = \sum_{i \mathop = 0}^\infty \frac 1 {i!}$ from the definition of Euler's number.

Multiplying both sides by $n!$, observe that:

$\displaystyle \frac m n n! = n! \sum_{i \mathop = 0}^\infty \frac 1 {i!} = \left({\frac{n!}{0!} + \frac{n!}{1!} + \frac{n!}{2!} + \cdots + \frac {n!}{n!}}\right) + \left({\frac{n!}{\left({n + 1}\right)!} + \frac{n!}{\left({n + 2}\right)!} + \frac{n!}{\left({n + 3}\right)!} + \cdots}\right)$

Observe that the quantity on the left must be an integer, as it is composed entirely of sums and differences of integral components.

It must be positive, as it is equal to:
 * $\displaystyle \frac 1 {\left({n + 1}\right)} + \frac 1 {\left({n + 1}\right) \left({n + 2}\right)} + \frac 1 {\left({n + 1}\right) \left({n + 2}\right) \left({n + 3}\right)} + \cdots$

which is strictly positive.

Thus:
 * $\displaystyle m \left({n - 1}\right)! - \left({\frac{n!}{0!} + \frac{n!}{1!} + \frac{n!}{2!} + \cdots + \frac{n!}{n!}}\right)$

must be a positive integer less than $1$.

From this contradiction it follows that $e$ must be irrational.