# Is this a valid proof that $e$ is irrational?

$$e=\sum_{n=0}^\infty\frac{1}{n!}$$

and assume for contradiction that $$e=\frac pq$$. Then we have that $$p=qe$$, where $$p$$ is an integer. So then $$q!e$$ is also an integer number (specifically $$p(q-1)!$$). This means that

$$q!e=q!\sum_{n=0}^\infty\frac{1}{n!}=\sum_{n=0}^\infty\frac{q!}{n!}\in\mathbb{Z}.$$

Now, we can split this sum into two parts:

$$q!e=\sum_{n=0}^q \frac{q!}{n!}+\sum_{n=q+1}^\infty\frac{q!}{n!},$$

and note that each term in the first sum is an integer (because $$n!|q!$$ when $$n\leq q$$). On the other hand, in the second sum,

$$\sum_{n=q+1}^\infty\frac{q!}{n!}=\sum_{n=q+1}^\infty\frac{q!}{q!(q+1)\cdots(q+n)}\leq\sum_{n=q+1}^\infty\frac{1}{(q+1)^n}<1$$

where there are $$n$$ terms in $$q!(q+1)\cdots(q+n)$$. This gives us our contradiction. Since the first sum in $$q!e$$ are all integers and the second sum is positive and less than 1, $$q!e\notin\mathbb{Z}$$ and thus $$e$$ is irrational.

I can't specifically say anything that's wrong with it, but I still don't feel good about it. In particular, I'm skeptical about rewriting $$n!=q!(q+1)\cdots(q+n)$$ when taking a sum to infinity, as well as the inequality, but I can't specifically say what might be wrong about it.

You should have written $$\sum_{n=q+1}^\infty \frac{q!}{n!} = \sum_{n=1}^\infty \frac{q!}{q!(q+1)\cdots(q+n)} \le \sum_{n=1}^\infty \frac{1}{(q+1)^n} = \frac{1/(q+1)}{1 - 1/(q+1)} = \frac{1}{q} < 1$$ since $$q > 1$$. Note the index of summation shifts. The argument you made is sound.