Prove $\lim \limits_{n \to \infty}\frac{n!}{n^n}=0$ Prove $\lim \limits_{n \to \infty}\frac{n!}{n^n}=0$.  I have tried this several ways.  I tried using the ratio test, I tried to expand it... I'm not really sure where to go.
When I expand, I get $$\frac{n(n-1)(n-2)(n-3).....}{nnnnnn......}$$ I don't know if trying to expand this is helpful, but I'm just not sure where to go with this problem... Any help would be appreciated.
 A: You've got the right idea. Note that
$$\frac{n!}{n^n}=\frac{n}{n}\frac{n-1}{n}\cdots\frac{1}{n}$$
Every term in this product is at most $1$ so we have that
$$\frac{n!}{n^n}=\frac{n}{n}\frac{n-1}{n}\cdots\frac{1}{n}\leq\frac{1}{n}$$
A: Here is a combinatorial proof. The numerator is the number of bijections from $[n]\to[n]$; the denominator is the number of functions from $[n]\to [n]$.  For each bijection $\sigma$, we can define $n$ different functions $\tau_1,\dots,\tau_n$ s.t. $\tau_i|_{[n-1]}=\sigma$ and $\tau_i(n)=i$. So the expression in the limit is $\leq 1/n$.
A: The ratio test works:
$$\frac{\frac{(n+1)!}{(n+1)^{n+1}}}{\frac{n!}{n^{n}}}=\frac{(n+1)!}{(n+1)^{n+1}}\frac{n^n}{n!}=\left(\frac{n}{n+1}\right)^n=\frac{1}{\left(1+\frac{1}{n}\right)^n}\overset{n\to\infty}{\longrightarrow} \frac{1}{e}<1.$$
A: Note that since $\sum_{k=1}^{n-1}\log(k)\le \int_1^{n} \log(x)\,dx$, we have
$$\begin{align}
(n-1)!&=e^{\log((n-1)!)}\\\\
&=e^{\sum_{k=1}^{n-1}\log(k)}\\\\
&\le e^{\int_1^{n} \log(x)\,dx}\\\\
&=n^ne^{-(n+1)}
\end{align}$$
Therefore, we see that 
$$0\le \frac{n!}{n^n}\le ne^{-(n+1)}$$
whereupon application of the squeeze theorem yields the coveted limit
$$\lim_{n\to \infty}\frac{n!}{n^n}=0$$
And we are done!
