Induction: $\frac{n!}{x(x+1)\cdots(x+n)} = \binom{n}{0}\frac{1}{x}-\binom{n}{1}\frac{1}{x+1}+\cdots+(-1)^n\binom{n}{n}\frac{1}{x+n}$ $$\frac{n!}{x(x+1)\cdots(x+n)} = \binom{n}{0}\frac{1}{x}-\binom{n}{1}\frac{1}{x+1}+\cdots+(-1)^n\binom{n}{n}\frac{1}{x+n}, \quad \text{for } x \not \in \{0,-1,-2,\dots,-n\}$$
Can somebody please help me with this?
I tried to multiply both sides with $x(x+1)\cdots(x+n)$.
$$n! = \sum_{k=0}^n (-1)^k\binom{n}{k}\frac{x(x+1)\cdots(x+n)}{x+k}$$
Then to divide both sides with $n!$.
$$1 = \sum_{k=0}^{n}\frac{(-1)^k}{k!(n-k)!}\frac{x(x+1)\cdots(x+n)}{x+k}$$
And then, assume it holds for $n$, and prove for $n+1$:
$$\sum_{k=0}^{n+1}\frac{(-1)^k}{k!(n+1-k)!}\frac{x(x+1)\cdots(x+n+1)}{x+k} = \\ \sum_{k=0}^{n}\frac{(-1)^k}{k!(n-k)!}\frac{x(x+1)\cdots(x+n)}{x+k}\frac{x+n+1}{n+1-k} + \\
\frac{(-1)^{n+1}}{(n+1)!}x(x+1)\cdots(x+n)$$
And this $\frac{x+n+1}{n+1-k}$ term is keeping me from applying the induction hypothesis.
EDIT: @Daniel's hint was pretty useful, thanks. After doing that, it boils down to this:
$$\sum_{k=0}^{n+1}\frac{(-1)^k}{k!(n+1-k)!}\frac{x(x+1)\cdots(x+n+1)}{x+k} = 1 + x(x+1)\cdots(x+n)\sum_{k=0}^{n+1}\frac{(-1)^k}{k!(n+1-k)!}$$
So, proving that $\sum_{k=0}^{n+1}\frac{(-1)^k}{k!(n+1-k)!} = 0$ should do it.

Hint.  Use that fact that the result is an identity for all $x$ (with the stated exceptions).  So, assume the result is true for a specific $n$.  Then we can say
$$\frac{n!}{x(x+1)\cdots(x+n)} = \binom{n}{0}\frac{1}{x}-\binom{n}{1}\frac{1}{x+1}+\cdots+(-1)^n\binom{n}{n}\frac{1}{x+n}\ ,$$
and also
$$\eqalign{&\frac{n!}{(x+1)(x+2)\cdots(x+n+1)}\cr
  &\qquad\qquad{}= \binom{n}{0}\frac{1}{x+1}-\binom{n}{1}\frac{1}{x+2}+\cdots+(-1)^n\binom{n}{n}\frac{1}{x+n+1}\ .\cr}$$
Now write down the RHS for $n+1$, simplify and use the above two equations.  You should find it just drops out.
