calculus - Limit of the sequence $frac {a^n} {n!}$

I need to prove that $$\lim_{n \rightarrow \infty} \frac {a^n} {n!}=0$$

I have no condition over $a$, just that is a real number. I thought of using L'Hôpital, but it's way too complicated for something that it should be simpler. Same goes to the epsilon proof, and I´m runnning out of options of what to do with it.

Thanks!

PD: I could also use a hint on how to solve $$\lim_{n \rightarrow \infty} \frac {n^n} {n!}=0$$

Answer

Compare it with the following geometric sequence:
$b_n=(\frac{a^m}{m!})(\frac{a}{m+1})^n,$ where $m$ is the smallest positive integer such that $m+1\gt a.$
Notice that $a_{n+m}\le b_n$ so that $\lim a_n=\lim b_n=0.$
Hope this helps.