How to show, that the number-sequence $a_n:=\frac{n^2}{n!}$ is bounded and has monotonicity? 
How to show, that the number-sequence $a_n:=\frac{n^2}{n!}$ with $(a_n)_{n\in\mathbb{N}}$ is bounded and has monotonicity?

Boundedness: I observed, that $a_1<a_2>a_3>a_4>a_5>\dots$ That's why I assume, that there is a bound $2=K\geq \mid a_n \mid$ with $K\in\mathbb{N}$. How do I show, that  $a_n$ is actually bounded at $K=2$?
Monotonicity: For $n\geq2$, we show
$\begin{align}
a_n&\geq a_{n+1}\\
\frac{n^2}{n!}&\geq \frac{(n+1)^2}{(n+1)!}\\
n^2(n+1)!&\geq (n+1)^2\cdot n!\\
(n+1)!&\geq n!
\end{align}$
For $n\leq 2$ we show
$\begin{align}
a_n&\leq a_{n+1}\\
\frac{n^2}{n!}&\leq \frac{(n+1)^2}{(n+1)!}\\
n^2(n+1)!&\leq (n+1)^2\cdot n!\\
n^2&\leq (n+1)^2
\end{align}$
but that's not really a proof. I don't know how to express formally, that $n!$ increases faster, than $n^2$ for $n\geq 2$ and the other way around for $n\leq 2$.
(Because of the fact, that, $n!$ increases faster, than $n^2$ for $n\geq 2$ and the other way around for $n\leq 2$ it's only important to look at $(n+1)!\geq n!$ or $n^2\leq (n+1)^2$  in the first place.)
 A: For $n \geq 4$ we have that $n^2 \leq n!$
Indeed it is true for $n=4$
Now assume that $m^2 \leq m!$ for some $m \in \Bbb{N}$
Then $(m+1)^2 \leq 4m^2 \leq 4(m!) \leq (m+1)m!$ since $m\geq 4$
For $n=1$ the value of the sequence is $1$
For $n=2$ the value is $2$
For $n=3$ the value is $\frac{9}{6}<2$
So the sequence is bounded by $K=2$.
A: For $n\ge 2$ you already have a proof, then for $n\le 2$ just consider $n=1$ and $n=2$ and this complete the proof by exhaustion.
A: Your proof is correct and  for $n\leq 2$ you can simply say that $a_1=1<a_2=2$.
With a little effort we can show more.
For $n\geq 2$: $0<a_{n+1}<a_n$ because
$$\frac{a_{n+1}}{a_n}=\frac{(n+1)^2}{(n+1)!}\cdot \frac{n!}{n^2}
=\frac{n+1}{n^2}=\frac{1}{n}+\frac{1}{n^2}\leq \frac{1}{2}+\frac{1}{4}=\frac{3}{4}<1.$$
Since $a_1=1<a_2=2$, it follows that $(a_n)_{n\geq 1}\in (0,2]$ and the maximum value is just $a_2=2$. 
By the above estimate we have also that for $n\geq 2$
$$0<a_{n+1}\leq \frac{3}{4}a_n\leq \left(\frac{3}{4}\right)^{2}a_{n-1}\leq \dots
\leq \left(\frac{3}{4}\right)^{n-1}a_{2}\to 0$$
and therefore $a_n=\frac{n^2}{n!}\to 0$.
