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.
Your guess is correct, and your problem can indeed be tackled by the Stirling's approximation:
$$\frac{r^n n!}{n^n} \sim \sqrt{2\pi n}\,\frac{r^n n^n e^{-n}}{n^n}=\sqrt{2\pi n}\,\left(\frac{r}{e}\right)^n.$$
Focusing on $r>0$, this limit converges if $r<e$ and diverges if $r\geq e$.
In fact, we may borrow the idea from the Ratio Test and come up with a simpler argument. If $a_n$ denotes the expression inside of the limit, then
$$\frac{a_{n+1}}{a_n}=\frac{r}{(1+\frac{1}{n})^n}.$$
Now by noting that $(1+\frac{1}{n})^n \uparrow e$ as $n\to\infty$,
$$ \frac{a_{n+1}}{a_n} \geq \frac{r}{e} \qquad\text{and}\qquad \lim_{n\to\infty} \frac{a_{n+1}}{a_n} = \frac{r}{e}. $$
This can be used to prove that $a_n \to 0$ if $0 < r < e$ and $a_n \not\to 0$ if $r \geq e$.