Easy proof, that $\rm e\notin \mathbb Q$

(Too long for a comment.) This is probably the way you did it, but it doesn’t seem very long to me; where did it bog down?

From the binomial theorem we have

$$a_n = \left(1 + \frac1n \right)^n = \sum\limits_{k=0}^n \binom{n}{k}n^{-k} = \sum\limits_{k=0}^n \frac{n^{\underline k}}{k! n^k} \le \sum\limits_{k=0}^n \frac{1}{k!} = b_n.$$

For the other direction let $m>1$ be an integer; then

$$a_{mn} = \sum\limits_{k=0}^{mn}\frac{(mn)^{\underline{k}}}{k!(mn)^k} \ge \sum\limits_{k=0}^n\frac{(mn)^{\underline{k}}}{k!(mn)^k} \ge \left(\frac{(mn)^{\underline n}}{(mn)^n} \right) b_n \ge \left(\frac{m-1}{m} \right)^n b_n,$$

and $\displaystyle \lim\limits_{m\to\infty} \left(\frac{m-1}{m}\right)^n = 1$, so $b_n \le e$.