Proof for convergence of a given progression $a_n := n^n / n!$

HINT: $$\begin{align*} \frac{n^n}{n!}&=\frac{n}{n}\cdot\frac{n}{n-1}\cdot\frac{n}{n-2}\cdot\dots\cdot\frac{n}2\cdot\frac{n}1\\ &=\left(\frac{n}{n-1}\cdot\frac{n}{n-2}\cdot\dots\cdot\frac{n}2\right)n \end{align*}$$

If you know that $$\lim_{n\to\infty}\left(1+\frac1n\right)^n = e\;,$$ you can also look at the ratio of consecutive terms to see that the sequence grows almost exponentially:

$$\begin{align*} \frac{a_{n+1}}{a_n} &=\frac{\frac{(n+1)^{n+1}}{(n+1)!}}{\frac{n^n}{n!}}\\ &=\frac{(n+1)^{n+1}}{n^n}\cdot\frac{n!}{(n+1)!}\\ &=\frac{(n+1)^n(n+1)}{n^n}\cdot\frac1{n+1}\\ &=\frac{(n+1)^n}{n^n}\\ &=\left(\frac{n+1}n\right)^n\\ &=\left(1+\frac1n\right)^n \end{align*}$$

Can you show $n! \le n^{n-1}$? Then you'd have $n^n/n! \ge n^n/n^{n-1} = n$, and $\lim_{n \to \infty} n = \infty$, so $\lim_{n \to \infty} n^n/n! = \infty$ as well.

Of course this is a very crude bound, but $n^n$ is so much larger than $n!$ that you don't need to be careful.