Partial fraction expansion of $\frac{1}{x(x+1)(x+2)\cdots(x+n)}$

Since every $k,\; k=-n,\ldots, 0$ is a simple pole of the given fraction then its decomposition take the form

$$\frac{1}{x(x+1)(x+2)...(x+n)}=\sum_{k=0}^n\frac{a_k}{x+k}$$ and we have $$a_k=\lim_{x \to -k}\sum_{i=0}^n\frac{a_i(x+k)}{x+i} = \lim_{x \to -k} (x+k)\sum_{i=0}^n\frac{a_i}{x+i}$$ $$= \lim_{x \to -k} \frac{x+k}{x(x+1)(x+2)...(x+n)}=\frac{1}{-k(-k+1)(-k+2)...(-k+n)}=\frac{(-1)^k}{k!(n-k)!}$$ so yes it's true that $$\frac{n!}{x(x+1)(x+2)...(x+n)}=\sum_{k=0}^n\frac{(-1)^k{n\choose k}}{x+k}$$

Notice that: $$ \frac{1}{x(x+1)\cdots(x+n)} = \frac{1}{n} \frac{(x+n) - x}{x(x+1)\cdots(x+n)} = \frac{1}{n} \left(\frac{1}{x(x+1)\cdots(x+n-1)} - \frac{1}{(x+1)\cdots(x+n)}\right) $$

This gives you a recursion formula for partial fraction expansion. You can use it to check your conjecture, and prove it by induction if it's indeed true.

See also this answer: Series of inverses of binomial coefficients for another approach.