What's the limit of the sequence $\lim\limits_{n \to\infty} \frac{n!}{n^n}$? $$\lim_{n \to\infty} \frac{n!}{n^n}$$
I have a question: is it valid to use Stirling's Formula to prove convergence of the sequence?

You can prove that
$$
n! < \left( \frac{n+1}{2} \right)^n.
$$
Now observe that
$$
0 \leq \lim_{n\to\infty} \frac{n!}{n^n} <\lim_{n\to\infty} \frac{\left(\frac{n+1}{2}\right)^n }{n^n} = \lim_{n\to\infty} \frac{1}{2^n}\cdot\frac{(n+1)^n}{n^n}.
$$
We know that $1/2^n\to 0$ as $n\to\infty$. If you know that $[(n+1)/n]^n\to e$ as $n\to\infty$, then you're done... the limit is zero!
Definitely not as nice as anon's solution, but a different approach nontheless.
