Get the right "upper" Sequence I want to find out the limit for the sequence $\frac{n!}{n^n}$ 
using the squeeze theorem.
My idea was:
$\frac{n}{n^n} \leq \frac{n!}{n^n} \leq \frac{n!+1}{n^n} $
So the limits of the smaller sequence and the bigger sequence are 0, thus
the limit of the original sequence is 0.
But is the bigger sequence $\frac{n!+1}{n^n}$ a good choice? Or is there
a more obvious sequence to choose? 

For the left one, I would personally even choose $n! \geq 0$. 
For the right one: Use the AM-GM inequality:
$$\frac{2n}{3}>\frac{(n+1)}{2}=\frac{\frac{1}{2}n(n+1)}{n} > \sqrt[n]{n!}$$
Therefore $n!< \left(\frac{2}{3}n\right)^n$. I'm sure you can finish it now. 

A more basic argument: 
A half of the terms are smaller than $n/2$, the other half is smaller than $n$, so $$n! \leq \left(\frac{1}{2}n\right)^{n/2}n^{n/2}=\frac{n^n}{2^{n/2}}$$
