Possible Duplicate:
Prove that $\lim \limits_{n \to \infty} \frac{x^n}{n!} = 0$, $x \in \Bbb R$.
I'm not sure how to go about solving this. Right now I'm trying to use the squeeze theorem. Notice $\frac{1}{n^n} <\frac{2^n}{n!} < \frac{2^n}{3^n}$. If I can prove the limit of the lower and upper sequences $= 0$, then by the squeeze theorem the limit of $\frac{2^n}{n}= 0$. However I dont know how to prove the limit of $\frac{1}{n^n} = 0$ so I'm stuck, perhaps there is a better way to solve this problem than the squeeze theorem?
Thanks.