In the identity,
$\displaystyle \frac{n!}{x(x+1)(x+2).....(x+n)} = \sum_{k=0}^n \frac{A_{k}}{x+k}$
prove that $\displaystyle A_{k}=(-1)^{k}\binom{n}{k}$
Thanks in advance
What does it mean to prove something "in the identity"? Do you mean to prove that the first identity implies the second one? That would be surprising because I would expect one can choose $\displaystyle A_k$ is many ways to satisfy the first formula.