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? 
 A: $$ \frac{n!}{n^n} =\frac{1}{n}\cdot\frac{2}{n}\cdots \frac{n-1}{n}\frac{n}{n}\le \frac{1}{n}.$$ That's because all the factors to the right of $1/n$ are smaller than or equal to $1$. We have equality when $n=1$, with $1/n$ being the only factor, and when $n=2$, with $1/n$ being the only factor besides $n/n=1$.
A: $$\frac {n!}{n^n} =\frac {n \times (n-1)\times \dots \times 2 \times 1}{n^n}\le \frac {\overbrace {n\times n \times \dots \times n}^{{n-1}\text {times}}\times 1}{n^n}=\frac {n^{n-1}}{n^n}=\frac1n$$
A: I think you are asking how large or small is $\frac{n!}{n^n}$, for large $n$.  
A rather clear answer is given by Stirling's formula that estimates $n!$.  The ratio of interest grows exponentially, at a rate of about $e^n$.
A: 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 $$n! \leq \left(\frac{1}{2}n\right)^{n/2}n^{n/2}=\frac{n^n}{2^{n/2}}$$
