Proof that $\lim\limits_{n\to \infty}\frac{n!}{n^n}$ is $0$

Question

As stated in the title, I'm trying to prove that the sequence $\{a_n\}$ defined by:

$$a_n=\frac{n!}{n^n}$$

Converges to zero. That is: $$\lim_{n\to \infty}\frac{n!}{n^n}=0$$

For example, I was thinking that for a sufficiently big $n$: $$a_n=\frac{n}{n}\cdot\frac{n-1}{n}\cdot\frac{n-2}{n}\cdots\frac{1}{n}= \frac{n-1}{n}\cdot\frac{n-2}{n}\cdots\frac{1}{n}$$ Then these $n-2$ factors $$\frac{n-1}{n}, \frac{n-2}{n}, \frac{n-3}{n},\cdots,\frac{2}{n}$$ are all smaller (or equal) than the first, then: $$a_n\le\Bigl(\frac{n-1}{n}\Bigr)^{n-2}\Bigl(\frac{1}{n}\Bigr)$$ Now, since $\frac{n-1}{n}\le1$, we can claim that: $$a_n\le\frac{1}{n}$$ which would instantly proof what I wanted by squeezing the sequence between $0$.
I'm not entirely convinced by this argument,and I have the feeling that there must be some mistake or something that I'm taking for granted. I'd love some feedback or even some other proofs for this.
Preferably, using only algebraic manipulations or calculus results.

Answer 1

It all looks good to me. It's difficult to address your concerns when I have no specific knowledge as to what's bothering you. Maybe presenting your proof in one line might help?

$$a_n = \frac{n}{n} \cdot \frac{n-1}{n} \cdot \ldots \cdot \frac{2}{n} \cdot \frac{1}{n} \le 1 \cdot 1 \cdot \ldots \cdot 1 \cdot \frac{1}{n} = \frac{1}{n} \to 0.$$

Perhaps your concern regards the expanding product, the number of terms of which increases with $n$? Maybe you're connecting it with the somewhat paradoxical nature of Riemann sums, adding $0$ "infinity times" to get a non-zero number? This isn't applicable here. Your proof shows that, at each step (when there's a finite number of terms), the total result of the product is bounded above by $\frac{1}{n}$. There's nothing wrong with the proof.

As for an alternate proof, you can show this is to show the infinite series $\sum_n a_n$ converges, by way of the ratio test. Consider, $$\frac{a_{n+1}}{a_n} = \frac{(n + 1)! n^n}{n! (n + 1)^{n+1}} = \frac{n^n}{(n+1)^n} = \left(1 + \frac{1}{n}\right)^{-n} \to \frac{1}{e} < 1.$$ By the ratio test $\sum_n a_n$ converges, and hence by the divergence test, $a_n \to 0$.