Partial answer
I knew the following, i hope you could take advantadge from this as well
Since $e^{n} = \sum\limits_{k=0}^{\infty} \frac{n^{k}}{k!}$ it's a non negative series containing the factor $\frac{n^{n}}{n!}$ for sure we obtain that $$e^{n} \geq \frac{n^{n}}{n!}$$
Which give us a first inequality $$n^{n} \geq n! \geq n^{n} \cdot e^{-n}$$
Fortunately we can find better ones,one of whom is given by the Mean Value Theorem with $f(x) = x\cdot log(x) \hspace{0.2cm}$:
$$f(k+1)-f(k) = (k+1)\cdot \log(k+1) - k\log(k)$$
$$(k+1)\log(k+1) - k\cdot \log(k) \geq (\log(k) +1)(k+1-k)$$
Summing for per $k=1,2,\cdots,n-1$ (Which give us a telescopic sum) we get :
$$n\cdot \log(n) \geq \sum\limits_{k=1}^{n-1}\log(k)+(n-1)$$ Adding $log(n)$ on both sides we reach : $$(n+1)\log(n) \geq \log(n!) + (n-1)$$
Taking the exponential $$n! \leq n^{n+1} \cdot e^{-n+1}$$
And we have just improved the first inequality with $$\left(\frac{n}{e}\right)^n \leq n! \leq e^2 \left(\frac{n}{e}\right)^{n+1}$$
Best Answer
Let $a_n =\displaystyle \frac{2^n n!}{n^n}.$ Note that $a_6 = 80/81 < 1.$ We also have $$ \frac{a_{n+1}}{a_n} = \frac{2^{n+1} (n+1)! }{(n+1)^{n+1}} \cdot \frac{n^n}{2^n n!} = 2 \left(\frac{n}{n+1}\right)^n < 1.$$ The sequence $$x_n = \left(1- \frac{1}{n+1} \right)^n$$ is monotonically decreasing to $1/e.$ Since $e>2$, $a_{n+1}/a_n < 1$ so $(a_n)$ is a monotonically decreasing sequence. Thus the first inequality holds.
By considering Taylor series, $\displaystyle e^x \geq \frac{x^n}{n!}$ for all $x\geq 0,$ and $n\in \mathbb{N}.$ In particular, for $x=n$ this yields $$ n! \geq \left( \frac{n}{e} \right)^n $$ and this is stronger than the second inequality.
We could have used the same proof method for the second inequality as we did for the first: Let $b_n= \displaystyle \frac{3^n n!}{n^n}.$ Then $b_6 = 45/4 > 1.$ Also, $$ \frac{b_{n+1}}{b_n} = \frac{3^{n+1} (n+1)! }{(n+1)^{n+1}} \cdot \frac{n^n}{3^n n!} = 3 \left(\frac{n}{n+1}\right)^n $$ and this is greater than $1$ since $e<3.$
What we have just done suggests we can prove the following: If $a,b$ are positive real numbers such that $a<e<b$ then $$ \left(\frac{n}{a} \right)^n > n! > \left( \frac{n}{b} \right)^n $$ holds for sufficiently large $n$. This is turn suggests something stronger about the growth of the factorial function: $n!$ is asymptotically equal to $(n/e)^n$ to within at most a sub-exponential factor.