Prove $2^n \in \mathcal{O}(n!)$

Question

Can someone help me with this? So far I have:

$2^n < c * n!$ for all $n \ge k$, where $c > 0$ and $k > 0$

$\frac {(2^n)}{n!} < c$ for all $n \ge k$

I'm not sure where to proceed.

Answer 1

If $n\ge3$,$$\frac{2^n}{n!}=\frac{8\cdot 2^{n-3}}{6\prod_{k=4}^nk}\le\frac{2^n}{6\cdot 4^{n-3}}=\frac{32}{3}2^{-n}\in o(1).$$