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!

As you said you need to check if there is a constant $C>0$ such that $n! \le C 2^n$ for all $n$ (or at least all sufficiently large $n$, but it does not matter much).
Note it is crucial that $C$ does not depend on $n$. 
In this case it turns out there is no such $C$. There are various ways to see this. One of them could go like this. 
Assume there is such a $C$. Then 
$n! \le C 2^n$ for all sufficiently large $n$. 
Then $\prod_{i=1}^n \frac{i}{2} \le C$ for all sufficiently large $n$.
Yet for $i  \neq 1$ we fave $\frac{i}{2}\ge 1$, so that $\prod_{i=1}^n \frac{i}{2} \ge \frac{1}{2} \frac{n}{2}$ (for $n \ge 2$). 
Thus we get $\frac{n}{4} \le C$ for  all sufficiently large $n$, which is absurd.
