From the Taylor series for $e^x$ we know the following equation: \begin{equation} e=\sum_{k=0}^{\infty}\frac{1}{k!}. \end{equation}Now let us assume that $e$ is rational. This would mean there are two natural numbers $a$ and $b$ such that: $$e=\frac{a}{b}.$$ This yields: $$b!e\in\mathbb{N}.$$ Now we can write $e$ using (1): $$b!e=b!\sum_{k=0}^{\infty}\frac{1}{k!}.$$ This can also be written: $$b!e=\sum_{k=0}^{b}\frac{b!}{k!}+\sum_{k=b+1}^{\infty}\frac{b!}{k!}.$$ The first sum is obviously a natural number, and thus $$\sum_{k=b+1}^{\infty}\frac{b!}{k!}$$ must also be natural. Now we see: $$\sum_{k=b+1}^{\infty}\frac{b!}{k!}=\frac{1}{b+1}+\frac{1}{(b+1)(b+2)}+...< \sum_{k=1}^{\infty}\left(\frac{1}{b+1}\right)^k=\frac{1}{b}.$$ Since $\frac{1}{b}\leq 1$ we conclude: $$0<\sum_{k=b+1}^{\infty}\frac{b!}{k!}<1.$$ We have also seen that this is an integer, but there is no integer between 0 and 1. So there cannot exist two natural numbers $a$ and $b$ such that $e=\frac{a}{b}$ so $e$ is irrational.