Determining the value to which the sequence $a_n=\frac{n!}{n^n}$ converges. How can it be deduced that the sequence $a_n=\dfrac{n!}{n^n}$ converges to $0$? I can reasonably infer this to be true, because I see the pattern as $n$ approaches larger values, but I am unsure of how to to evaluate $\lim a_n=L$ to show that $L=0$.
 A: For $n > 1$,
$$\frac{n!}{n^n} = \prod_\limits{i = 1}^{n}\frac{i}{n} = \frac{1}{n} \prod_\limits{i = 2}^{n}\frac{i}{n} \leq \frac{1}{n}$$
It is bounded by a sequence $\frac{1}{n}$ that converges to zero, so it must converge to zero to.
A: HINT:
$$\begin{align}
\frac{a_{n+1}}{a_n}&=\frac{(n+1)!}{(n+1)^{n+1}}\frac{n^n}{n!}\\\\
&=\frac{1}{\left(1+\frac1n\right)^n}\\\\
&\to e^{-1}\\\\
&<1
\end{align}$$

Of course, Stirling's formula is the better way to go.  Although the OP mentioned that that topic has not yet been discussed, I thought that it might be instructive to show how it can be used here.  So, here we go ...
Stirling's formula is given by 
$$n!=\sqrt{2\pi n}\left(\frac{n}{e}\right)^n\left(1+O\left(\frac1n\right)\right)$$
Thus, we have
$$\frac{n!}{n^n}=\sqrt{2\pi n}e^{-n}\left(1+O\left(\frac1n\right)\right)$$
which clearly goes to zero as $n \to \infty$.  
It is also worth remarking that from Stirling's formula we obtain
$$\frac{a_{n+1}}{a_n}=e^{-1}\left(1+O\left(\frac1n\right)\right)$$
which goes to $e^{-1}$ as $n\to \infty$ as expected.
A: Write $\frac{n!}{n^n}$ as 
$$
\frac{n}{n}\frac{n-1}{n} \cdots \frac{1}{n}
$$
and as you can see by the last factor of this, $\frac{n!}{n^n} \leq \frac1n$ (because all the other factors are less than $1$).
Obviously we have $0 \leq \frac{n!}{n^n} \leq \frac 1n$ , so we know that it has to converge to $0$
