Prove by induction $\frac{n^n}{3^n}$\dfrac{n^n}{3^n}<n!<\dfrac{n^n}{2^n}$ 
The case $n!<\dfrac{n^n}{2^n}$ is easier. 

Another way this might be viewed is by rearranging the inequalities as
$$2^n \ < \ \frac{n^n}{n!}  \ \ \text{and} \ \ \frac{n^n}{n!} \ < \ 3^n , $$
the entirety of which holds for $ \ n \ = \ 6 \ , $ as already mentioned by user84413 .  The "induction step" for the ratio is then
$$\frac{(n+1)^{n+1}}{(n+1)!} \ = \ \frac{(n+1)}{(n+1)} \ \cdot \ \frac{(n+1)^{n}}{n!} \ = \  \left( \frac{n+1}{n} \right)^n \ \cdot \ \frac{n^{n}}{n!} \ . $$
We know that this first factor produces a number between 2 and 3 for $ \ n \ \ge \ 1 \ $ (in fact, it's a familiar statement that $ \ \lim_{n \rightarrow \infty} \ \left( \frac{n+1}{n} \right)^n  = \ e \ )^{*} \ , $
$^{*}$ indeed, some regard this as the defining equation for $ \ e \ $
so we can write
$$2^{n+1} \ = \ 2 \ \cdot \ 2^n  \ < \ 2 \ \cdot  \frac{n^n}{n!} \ \le \ \left( \frac{n+1}{n} \right)^n \ \cdot \ \frac{n^{n}}{n!}  \ = \ \frac{(n+1)^{n+1}}{(n+1)!} \  $$
[the equation being true for $ \ n = 1 \ $ ]
and
$$ \frac{(n+1)^{n+1}}{(n+1)!} \ = \ \left( \frac{n+1}{n} \right)^n \ \cdot \ \frac{n^{n}}{n!}  \ < \ 3\ \cdot  \frac{n^{n}}{n!}  \ < \ 3 \ \cdot \ 3^n \  =  \ 3^{n+1} \ . $$
Thus, for $ \ n \ \ge \ 6 \ , $ we find  $ \ 2^n \ < \ \frac{n^n}{n!}  \  < \ 3^n \ , \ $ equivalent to the original inequality 
$$ \ \frac{n^n}{3^n} \ < \ n!  \  < \ \frac{n^n}{2^n} \ . \ $$ 
$$\\$$
Note that in so doing, we have also proven that $ \ \left(\frac{1}{3}\right)^n  < \ \frac{n!}{n^n}  \  < \ \left(\frac{1}{2}\right)^n \ , \ $ and hence, by the "Squeeze Theorem", that $ \ \lim_{n \rightarrow \infty} \ \frac{n!}{n^n} \  = \ 0 \ . $  The induction work is similar to that required in applying the Ratio Test to demonstrate the absolute convergence of $ \ \sum_{n=1}^{\infty} \ \frac{n!}{n^n} \ $ (and so also the divergence of $ \ \sum_{n=1}^{\infty} \ \frac{n^n}{n!} \ ) \ . $
ADDENDUM:  It might seem, from the preceding discussion, that we ought to get something like an equation out of this by just using $ \ e \ $ ; but, in fact we find that $ \ \frac{n^n}{e^n}  < \ n!  \ . \ $  This touches on the Stirling approximation, which requires some additional non-linear factors in order to get closer to accurate values of $ \ n! \ $ for large $ \ n \ . $
