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? 

$$\frac {n!}{n^n} =\frac {n \times (n-1)\times \dots \times 2 \times 1}{n^n}\le \frac {\overbrace {n\times n \times \dots \times n}^{{n-1}\text {times}}\times 1}{n^n}=\frac {n^{n-1}}{n^n}=\frac1n$$
