How to check if $n!$ is $ O(2^n)$ How can I check if $n! \in O(2^n)$?
The definition of $f$ being $O(g)$ is  $f(n) \le c g (n)$, where $c>0$.
So it would mean $n!  \le  c  2^n$.
What is the clearest way to solve this? (As I am solving the past papers for my next examen.)
Thank you!

A simple proof without induction: suppose there exists a constant $C>0$ such that $n!\le 2^n$. Note $n!\ge2^{n-1}$ for all $n\ge 1$, hence for all $n>1$,
$$n=\frac{n!}{(n-1)!}\le \frac{C2^n}{2^{n-2}}=4C$$
from which we conclude that $\mathbf N$ is bounded from above…
