Given:
$\frac{n^n}{3^n} < n! < \frac{n^n}{2^n} \forall n \geq 6$
Using Sterling's formula we want to prove the following 2 inequalities:
$\frac{n^n}{3^n} < \sqrt{2\pi n}\left ( \frac{n}{e} \right )^n ---> [1] $
and
$\frac{n^n}{2^n}>\sqrt{2\pi n}\left ( \frac{n}{e} \right )^n ---> [2] $
For the first inequality suppose that the opposite is true; that is, suppose that:
$\frac{n^n}{3^n} \geq \sqrt{2\pi n}\left ( \frac{n}{e} \right )^n ---> [3] $
so, $n^n \geq \sqrt{2\pi n}\left ( \frac{3*n}{e} \right )^n$
since $3/e$ is about $1.10363832351$ we'd have:
so, $n^n \geq \sqrt{2\pi n}\left ( 1.01*n \right )^n$
The above is clearly not true for any value of $n \geq 6$. Hence the proposition in [3] is false and [1] is true.
The same idea can be used to prove iniquality [2]. Note that $2/e$ is approximately 0.73575888234.