How to show that $e$ is irrational by studying $\sum^{n}_{k=1}\frac{1}{k!}$ and $\sum^{n}_{k=0}\frac{1}{k} + \frac{1}{n \cdot n!}$?

Assume that $e=p/q$ with $p,q\in \mathbb{N}^+$ then for $n=q$, $$(qq!)\cdot a_q <(qq!)\cdot e <(qq!)\cdot b_q\implies qq!a_q<pq(q-1)!<qq!a_q+1.$$ which means that the integer $pq(q-1)!$ is strictly between the two consecutive integers $qq!a_q$ and $qq!a_q+1$. Contradiction.

P.S. $(qq!)\cdot a_q$ is an integer because $$(qq!)\cdot a_q=q\sum^{q}_{k=0}\frac{q!}{k!}= q\sum^{q-1}_{k=0}\prod_{k+1}^q j+q$$ which is a sum of integers.

Hint: Let $n=b$, and clear the denominators in the inequality $a_n<e<b_n$.