Is it true that $2^n$ is $O(n!)$? I had a similar problem to this saying:
Is it true that $n!$ is $O(2^n)$?
I got that to be false because if we look at the dominant power of $n!$ it results in $n^n$. So because the base numbers are not the same it is false.
Is it true that $2^n$ is $O(n!)$?
So likewise with the bases, this question should result in false, however it is true. Why? Is the approach I am taking to solve these questions wrong?

If $f(n)$ is $O(g(n))$, then there is a constant $C$ such that $f(n) \leq C g(n)$ eventually.
It turns out that $2^n$ is $O(n!)$. Can you find a constant $C$ and prove the inequality? Hint: Choose $C=2$ and try to prove $2^n \leq 2 n!$.
