Limits of function and squeeze theorem

Question

By first showing that

$\frac{2^n}{n!}<18(\frac{2}{3})^n \quad$

for all n is an element of real numbers, find

$\lim \limits_{n \to \infty} \frac{2^n}{n!}$

by squeeze theorem.

I'm already stuck trying to prove the first part...I managed to resolve the inequality to:

$\frac{2^n}{n!}<\frac{2^{n+1}}{3^{n-2}}$

Answer 1

$18(\frac{2}{3})^n = 18(\frac{2^n}{3^n})$ which, when compared to $(\frac{2^n}{n!})$ is always bigger. $3^n$ is smaller than $n!$ for $n>6$ and hence $(\frac{2^n}{3^n})>(\frac{2^n}{n!})$ here. Below that, the $18$ factor ensures $18(\frac{2^n}{3^n})>(\frac{2^n}{n!})$