So, I know that $n!<(\frac{n}{2})^n$ for $n\geq6$.
The natural question that arises is:
Is it true that for every $k \in \mathbb N$ there exists $n_0(k)$ such that $n!<(\frac {n}{k})^n$ for every $n\geq n_0(k)$?
I guess that the answer is yes and that there should be some elementary enough way to prove it but I do not have an idea on how to start the proof.
The answer is no. Stirling's formula tells us that
$n! \ge \sqrt{2\pi} \frac{n^{n+0.5}}{e^n}>(\frac{n}{e})^n>(\frac{n}{3})^n$.
So no such result is possible for $k\ge 3$.
Here is an approach avoiding Stirling's formula
$\ln(n!)=\sum_{k=1}^{n} \ln(k) \ge \int_1^n \ln(x)\, dx= x\ln{x}-x|_1^n=n\ln{n}-n+1.$
So $n!>e^{n\ln{n}-n}=(\frac{n}{e})^n$, as desired.