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: Multiply both side by $x+k$ and then put $x=-k.$
