Euler's Number is Irrational

Theorem:
Euler's constant $$e\,\!$$ is irrational.

Proof by Contradiction:
Assume that $$e\,\!$$ is rational. Then, there exist coprime integers $$m$$ and $$n$$ such that: $$\frac{m}{n} = e = \sum_{i=0}^{\infty}\frac{1}{i!}$$

Multiplying both sides by $$n!\,\!$$, observe that: $$\frac{m}{n}n! = n!\sum_{i=0}^{\infty}\frac{1}{i!} = \left (\frac{n!}{0!} + \frac{n!}{1!} + \frac{n!}{2!} + \ldots + \frac{n!}{n!} \right ) + \left ( \frac{n!}{(n+1)!} + \frac{n!}{(n+2)!} + \frac{n!}{(n+3)!} + \ldots \right )$$

Observe that the quantity on the left must be a positive integer, as it is composed entirely of sums of integral components and is equal to a sum of positive terms, but it must also be strictly less than 1, a contradiction, so $$e\,\!$$ must be irrational.

Q.E.D.