Need to prove that $A_k=(-1)^k {n \choose k}$

Question

In the identity $$ \frac{n!}{x(x+1)(x+2)\ldots(x+n)}=\sum_{k=0}^n{\frac{A_k}{x+k}}, $$ prove that $$ A_k=(-1)^k {n \choose k}\text{ .} $$

How should I proceed with this problem? A push in the right direction or stating the prerequisites for this problem will be appreciated.

Answer 1

If we multiply both sides by $(x+k) $ and in the resulted equality, we replace $x $ by $-k $, we find

$$0+0+...+A_k+0+0+...0=$$ $\frac {n!}{-k (1-k)... (k-1-k) . (k+1-k)(k+2-k)... (n-k) }=$

$$\frac {n!}{((-1)^k.k!)((n-k)!}= $$

$$(-1)^k\binom{n}{k}=A_k$$

This technic is used in partial fractions decomposition.