Question: Prove that $$\lim\_{n\to \infty}\frac{(-1)^n(n+1)!}{n^{n+1}}=0$$

I noticed that it can be expressed in the form of $n+1$ terms product but I have no idea with the upper bound. 

I have trouble with how to operate $(-1)^n$

Answer: 
Let $\displaystyle a\_n=\frac{(-1)^n(n+1)!}{n^{n+1}}$. Then, 
$$\left|\frac{a\_{n+1}}{a\_n}\right|=\frac{n^{n+1}}{(n+1)^{n+2}}(n+2)=\frac{1}{(1+\frac 1n)^n}\frac{n(n+2)}{(n+1)^2}\xrightarrow{n\to \infty}\frac{1}{e}<1.$$

Hence, $a\_n \to 0$ as $n\to \infty$.
$$\lim\_{n\to \infty}\left|(-1)^n\right| = \lim\_{n\to \infty}\left|-1\right|^n = \lim\_{n\to \infty}1^n = \lim\_{n\to \infty}1 = 1$$

$$\lim\_{n\to \infty}(-1)^n = 1$$

$$\lim\_{n\to \infty}\frac{(-1)^n(n+1)!}{n^{n+1}} = \lim\_{n\to \infty}\frac{(-1)^n}{\frac{n^{n+1}}{(n+1)!}} = \frac{\lim\_{n\to \infty}(-1)^n}{\lim\_{n\to \infty}\frac{n^{n+1}}{(n+1)!}} =\frac{\mathrm{finite}}{\infty} = 0$$