Limit of $n!/n^n$ as $n$ tends to infinity Using a calculator, I found that $n!$ grows substantially slower than $n^n$ as $n$ tends to infinity. I guess the limit should be $0$. But I don't know how to prove it. In my textbook a hint is given that: 
Set $a_n=n!/n^n$ Set $m=[n/2]$(floor function), then $a_n \le (1/2)^m\le(1/2)^{n/2}$. Then by comparing to the geometric progression, the sequence $a_n$ tends to $0$.
I have trouble proving the relationship $a_n \le (1/2)^m\le(1/2)^{n/2}$, (I tried to prove by considering separate cases, that is when $m$ is odd and when it is even) using induction gets me nowhere. Or is there other way to prove this limit? I made some search on web and used Stirling's approximation, but to no avail.
P/S: Although some said that my question is probably duplicate, the main point in my question is understanding and proving the relationship of the inequalities, which I had trouble understanding and was not addressed in the other suggested question(the sequence $(1/n)$ was used as comparison instead of $(1/2)^{n/2}$.) 

Hint: Use the $\{x_n\}$ be sequence of positive real numbers such that $L=\lim \frac {x_{n+1}}{x_n}$ exists. If $L\lt1$, then $\{x_n\}$ converges and $\lim x_n=0$. Now in your case $$\lim \frac {x_{n+1}}{x_n}=\lim \frac {(n+1)!\times n^n}{(n+1)^{n+1}\times n!}\\
=\lim \left[\frac{n}{n+1}\right]^n=\lim \frac {1}{\left(1+\frac{1}{n}\right)^n}=1/e\lt1$$
Now Using above statement we have, $$\lim \frac{n!}{n^n}=0$$
