Answer
converges to $0$
Work Step by Step
As we know that when $x \gt 0$ so, $\lim\limits_{n \to \infty} \sqrt[n] {n}=1$ and $\lim\limits_{n \to \infty} x^{1/n}=1$
Let $\lim\limits_{n \to \infty} a_n= \lim\limits_{n \to \infty} \dfrac{n!}{n^n}$
But sequence $a_{n}$ is both bounded and monotonic so, the sequence $a_{n}$ converges.
This implies that $\lim\limits_{n \to \infty} a_n= \lim\limits_{n \to \infty} \dfrac{n!}{n^n} \leq \lim\limits_{n \to \infty} (\dfrac{1}{n}) =0$
Thus, $\lim\limits_{n \to \infty} a_n=0$ and {$a_n$} converges to $0$.