I have proved by induction that the Statement above is true
Here is my proof:
Inductionbase: $\frac{2!}{2^2}=\frac{2}{4}$
Inductionstep:$\frac{(n+1)!}{(n+1)^{n+1}}=\frac{n!}{(n+1)^n}\frac{n+1}{n+1}\overset{(n+1>n)}{\leq}\frac{n!}{n^n}\overset{\text{IH}}{\leq}\frac{2}{n^2}$
But I still don't understand why my proof is valid, why it makes sense. For me it is just a Formula, can somebody explainme how someone came up with the idea that:
$$\frac{n!}{n^n}\leq\frac{2}{n^2},\forall n\geq 2$$
On
Your proof is invalid, because you have to show that
if $\dfrac{n!}{n^n}\le\dfrac{2}{n^2}$, then $\dfrac{(n+1)!}{(n+1)^{n+1}}\le\dfrac{2}{(n+1)^2}$
Now $$ \frac{(n+1)!}{(n+1)^{n}}=\frac{n!}{(n+1)^n}=\frac{n!}{n^n}\frac{n^n}{(n+1)^{n}}\le \dfrac{2}{n^2}\frac{n^n}{(n+1)^{n}} $$ by the induction hypothesis.
You can finish by proving that, for $n\ge2$, $$ \dfrac{2}{n^2}\frac{n^n}{(n+1)^{n}}\le\dfrac{2}{(n+1)^2} $$ that's equivalent to $$ \frac{n^n}{(n+1)^{n}}\le\dfrac{n^2}{(n+1)^2} $$ or, as well, to $$ \left(\frac{n}{n+1}\right)^{n-2}\le1 $$ which follows from $n/(n+1)<1$: if $0<x<1$, then $x^k<1$, for $k\ge1$ (easy induction).
Since the proof only holds for $n\ge2$, you need to prove two base steps, for $n=1$ and $n=2$. Since $$ \frac{1!}{1^1}\le\frac{2}{1^2} \qquad \frac{2!}{2^2}\le\frac{2}{2^2} $$ you're done.
I would write the inequality in the form $$n!\le 2\cdot n^{n-2}$$