Show that $ \lim_{n \rightarrow \infty} \frac{n!}{2^{n}} = \infty $ Show that $ \lim_{n \rightarrow \infty}  \frac{n!}{2^{n}} = \infty $
I know what happens intuitively.... 
$n!$ grows a lot faster than $2^{n}$ which implies that the limit goes to infinity, but that's not the focus here.
I'm asked to show this algebraically and use the definition for a limit of a sequence.
"Given an $\epsilon>0$ , how large must $n$ be in order for $\frac{n!}{2^{n}}$ to be greater than this $\epsilon$ ?"
My teacher recommends using an inequality to prove it but I'm feeling completely lost...

For $n\geq8$
$$\frac{n!}{2^n}=\frac{2\cdot3}{2^4}\frac{4\cdot5...(n-1)}{2^{n-4}}n>\frac{2\cdot3}{2^4}\frac{4^{n-4}}{2^{n-4}}n=\frac{3}{2^4}2^{n-4}n\geq n$$
So, $n>\max(8,\epsilon)$ works.
