Proof or refute $n^n = \Omega(n!)$ with the help of Stirling's approximation
Stirling's approximation states that
$$ n! \sim \sqrt{2\pi n} (n/e)^n. $$
This notation means that the ratio between the two sides tends to 1 as $n$ tends to infinity. For your purposes, we can simply write
$$ n! = \Theta(\sqrt{n} (n/e)^n). $$
Since $\sqrt{n} = o(e^n)$,
$$ n! = o(n^n). $$

If all you want to show $n! = O(n^n)$, then as Jukka mentions you can use the simple bound $n! \leq n^n$.