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 $(1/2)^{n/2}$.) 
 A: 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, $$\lim \frac{n!}{n^n}=0$$
A: Note that
$$0 < \frac{n!}{n^n} = \frac{n \times (n - 1) \times \cdots \times 1}{n \times n \times \cdots \times n} = 1 \times \frac{n - 1}{n} \times \cdots \times \frac{1}{n} \leq 1 \times 1 \times \cdots \times \frac{1}{n} = \frac{1}{n}.$$
The answer $0$ follows from the squeeze principle.
A: Hint: $a_n = \frac{n}{n} \frac{n-1}{n} \ldots \frac{3}{n} \frac{2}{n} \frac{1}{n}$. 
Each term in the product is at most $1$ (so the first n/2 terms are less than 1 each), and the last $n/2$ terms are less than $\frac{1}{2}$. 
So, $a_n \leq (1/2)^\frac{n}{2}$
A: One more way: consider $e^{\log \frac{n!}{n^n}} = e^{\sum_k \log n - n \log n}$. Take bounds on the sum:
$$
\int_{1}^{n} \log x dx < \sum_{k=1}^{n} \log k <1 + \int_{1}^{n} \log x dx
$$
and the result follows by the squeeze thm. 
