How to show that $\sum_{n=1}^\infty (-1)^n \frac{n^n}{n!}$ diverges? I know that this series diverges. 

$$\sum_{n=1}^\infty (-1)^n \frac{n^n}{n!}$$

From what I learned, as it's an alternating series, I have to prove that $\frac{n^n}{n!}$ descends and $\lim_{n\to \infty} \frac{n^n}{n!} = 0$.
However, I don't know what method I should use to break down $\frac{n^n}{n!}$.
Thanks for your help.
(Also it's my first time editing formally like this, so if there's any mistake, any correction is appreciated!)
 A: Since $n^n>n!$ as easily seen by induction you have that if you call with $a_n$ the general term of your series, it does not have limit $0$ as $n \to \infty$.Therefore the series is divergent.
A: HINT. Apply the Ratio Test carefully.
As a general note, some series which are alternating, are much easier shown to be convergent using the Root/Ratio Test (at least those which are absolutely convergent). 
A: Stirling's formula for $n!$ is $$n! \approx \sqrt{2\pi} \, n^{n+1/2} e^{-n}$$
So the $n^\textrm{th}$ term of your series is approximately
$$a_n \approx (-1)^n \frac{e^n}{\sqrt{2\pi n} }.$$
The root test gives $|a_n|^{1/n} \rightarrow e >1.$
So the series diverges.
A: A necessary condition for any series $\sum_{n=1}^\infty a_n$ to converge is that $\lim_{n\to\infty} |a_n| =0$.
In your case is $a_n = (-1)^n \frac{n^n}{n!}$ and it is easy to verify that this sequence does not converge. For $n >1$ you have
$$|a_n| = \frac{\overbrace{n \cdots n}^{\lfloor \frac n2 \rfloor\; factors} \cdot n \cdots n}{1 \cdots \lfloor \frac n2 \rfloor \cdot (\lfloor \frac n2 \rfloor +1) \cdots n} > 2^{\lfloor \frac n2 \rfloor}\stackrel{n\to \infty}{\longrightarrow}+\infty$$
So, your series does not converge.
