Proof for convergence of a given progression $a_n := n^n / n!$ "Examine whether the given progressions converge (eventually improperly) and determine their limits where applicable.
(a) $$(a_n)_{n\in\mathbb{N}}:=\frac{n^n}{n!}$$
[...]"
I am having problems getting this homework done as I have no clue about convergency at all. Through Mathematica I know that $\lim\limits_{n\to\infty}a_n=+\infty$ however I don't know how to proove it! During my research I found out that for sufficient big $n$ the inequality $x^n\leq n!\leq n^n$ for $x\in\mathbb{R}$ and $n\in\mathbb{N}$ is true. However this doesn't help me at all. Having a look at our lecture notes I developed the following statement: every $x^n$ with $|x| \geq 1$ diverges as $x^n$ is unbounded and therefore has the improper limit $\lim\limits_{n\to\infty}x^n=+\infty$. Furthermore it seems obvious to me that I need to express the terms in a simpler way, like fractions converging to 0 etc, but I don't know how. Neither some work with the $\varepsilon$-definition of limits in combination with the triangle inequality helped me out. Any suggestions or hints?
Information: This is "Analysis for computer scientist" and therefore we haven't "officially" learned any fancy tricks like L'Hôpital's rule etc. just basic properties of $\mathbb{R}$ and some inequalities as this is an introductory lecture.

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