Determine $\alpha$, where $$ \alpha \equiv \sup \left\{r \in \mathbb{R} : \lim_{n \to \infty} \frac{r^n n!}{n^n} = 0 \right\} $$
Some Observations
We clearly have $\alpha \geq 1$ since $$ \lim_{n \to \infty} \frac{n!}{n^n} = 0 $$ Moreover, Wolframalpha tells me $$ \lim_{n \to \infty} \frac{4^n n!}{n^n} = \infty $$ so $\alpha \leq 4$.
A basic estimate shows that $$ 0 \leq \frac{r^n n!}{n^n} \leq \frac{r^n}{2^n} \left(1 + \frac{1}{n} \right)^n $$ which may be some kind of clue.
This is reminiscent of Stirling's approximation, which says $$ \lim_{n \to \infty} \frac{e^n n!}{n^{n + \frac{1}{2}}} = \sqrt{2\pi} $$ But the $n + \frac{1}{2}$ in the exponent in the denominator prevents us from simply concluding that $\alpha = e$
Playing around in Wolframalpha, it seems like $\alpha = e$ is a good conjecture, but I haven't been able to confirm it. I also have no idea how to prove it.