e is irrational

From the Taylor series for $e^x$ we know the following equation:

$$e=\sum _{k=0}^{\mathrm{\infty }}\frac{1}{k!}.$$(1)

Now let us assume that $e$ is rational. This would 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):

$$b!e=b!\sum _{k=0}^{\mathrm{\infty }}\frac{1}{k!}.$$

This can also be written:

$$b!e=\sum _{k=0}^b\frac{b!}{k!}+\sum _{k=b+1}^{\mathrm{\infty }}\frac{b!}{k!}.$$

The first sum is obviously a natural number, and thus

$$\sum _{k=b+1}^{\mathrm{\infty }}\frac{b!}{k!}$$

must also be . Now we see:

Since $\frac{1}{b}\le 1$ we conclude:

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.