How to show that $A_k=(-1)^k\binom nk$? 
In the identity $$\frac{n!}{x(x+1)(x+2)\cdots(x+n)}=\sum_{k=0}^n\frac{A_k}{x+k},$$prove that $$A_k=(-1)^k\binom nk.$$

My try: The given identity implies $$\frac{1\cdot2\cdots n}{x(x+1)(x+2)\cdots(x+n)}=\frac{A_0}{x}+\frac{A_1}{x+1}+\dots+\frac{A_n}{x+n}.$$
Now putting $A_k=(-1)^k\binom nk,$$$\frac{1\cdot2\cdots n}{x(x+1)(x+2)\cdots(x+n)}=\frac1x-\frac{n}{x+1}+\dots+\frac{(-1)^n}{x+n}.$$ How to proceed further?

Hint:
The number $A_k$ is the residue of the meromorphic function $$\frac{n!}{x(x+1) \cdots (x+n)} $$ at the point $x=-k.$ Since it's a simple pole the residue is $$A_k=\frac{n!}{\frac{d}{dx}[x(x+1) \cdots (x+n)]} \bigg|_{x=-k}=\frac{n!}{(-k)(-k+1) \cdots (-k+n)}, $$where in the final product the term $(-k+k)$ is obviously omitted.
