I try to understand what I've overlooked, when I came up with this inequality:
First, we have this limit: $$\lim\limits_{n \to \infty} \sqrt[n]{\frac{n!}{n^n}} = \frac{1}{e}$$ Which gives, by the definition of limit and some simple transformations:
$\frac{1}{e} - \varepsilon < \sqrt[n]{\frac{n!}{n^n}} < \frac{1}{e} + \varepsilon$
$(\frac{n}{e} - n\varepsilon)^{n} < n! < (\frac{n}{e} + n\varepsilon)^{n}\quad\forall \varepsilon > 0$
Then, we have this well-known inequality (multiple proofs can be found on math.stackexchange): $$(\frac{n}{e})^{n} < n!$$ So we have: $$(\frac{n}{e})^{n} < n! < (\frac{n}{e} + n\varepsilon)^{n}$$ According to this inequality, we cannot make $\varepsilon$ arbitrary small, which contradicts the definition of limit. What am I missing here?
Actually yes, we can make $\epsilon$ arbitrarily small. Note that your argument is a limit argument, meaning it doesn't hold for every $n$. It only holds for all $n \geq N$, where $N$ depends on $\epsilon$.