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?

We will first show that the sequence $x_n = \frac{n!}{n^n}$ converges. To do this, we will show that the sequence is both monotonic and bounded.
Lemma 1: $x_n$ is monotonically decreasing.
Proof. We can see this with some simple algebra:
$$x_{n+1} = \frac{(n+1)!}{(n+1)^{n+1}} = \frac{n+1}{n+1}\frac{n!}{(n+1)^n} \frac{n^n}{n^n} = \frac{n!}{n^n} \frac{n^n}{(n+1)^n} = x_n \big(\frac{n}{n+1}\big)^n.$$
Since $\big(\frac{n}{n+1}\big)^n < 1$, then $x_{n+1} < x_n$.
Lemma 2: $x_n$ is bounded.
Proof. Straightforward to see that $n! \leq n^n$ and $n! \geq 0$. We obtain the bounds $0 \leq x_n \leq 1$, demonstrating that $x_n$ is bounded.
Together, these two lemmas along with the monotone convergence theorem proves that the sequence converges.
Theorem: $x_n \to 0$ as $n \to \infty$.
Proof. Since $x_n$ converges, then let $s = \lim_{n \to \infty} x_n$, where $s \in \mathbb{R}$. Recall the relation in Lemma 1:
$$x_{n+1} = x_n \big(\frac{n}{n+1}\big)^n = \frac{x_n}{(1+ \frac{1}{n})^n}.$$
Since $x_n \to s$, then so does $x_{n+1}$. Furthermore, a standard result is the limit $(1+ \frac{1}{n})^n \to e$. With these results, we have $\frac{x_n}{(1+ \frac{1}{n})^n} \to \frac{s}{e}$ and consequently
$$s = \frac{s}{e} \implies s(1 - e^{-1}) = 0$$
Since $1 \neq e^{-1}$, then this statement is satisfied if and only if $s = 0$ and that concludes the proof.
