Q:

Proof by induction, or without it if possible?

I was given a task to prove:
$$
\frac{1}{(x+1)(x+2)\ldots(x+n)}=\frac{1}{(n-1)!}\sum_{i=1}^n\binom{n-1}{i-1}\frac{(-1)^{i-1}}{x+i}
$$
I am almost 100% sure this is best solved by induction but to be honest, I'm scared to start using induction for such an expression on a test where I am limited with time. This reminds me of partial fraction decomposition but that binomial coefficient is disturbing me. I'm not sure how difficult this is to prove, but what I am looking for is a way to do this without induction but it seems impossible to do so without it. Induction in general bothers me. I agree it's a fine way of proving something but I would always chose another way around. Unfortunately, most problems I encountered can't be solved any other way. So before you hit the "report as not real question" button, I would like to ask for a HINT on how to solve this, with or without induction? 

A:

Since every $k,\; k=-n,\ldots, -1$ is a simple pole of the given fraction then its decomposition take the form
$$\frac{1}{(x+1)(x+2)...(x+n)}=\sum_{k=1}^n\frac{a_k}{x+k}$$
and we have
$$a_k=\lim_{x \to -k}\sum_{i=1}^n\frac{a_i(x+k)}{x+i} = \lim_{x \to -k} (x+k)\sum_{i=1}^n\frac{a_i}{x+i}
= \lim_{x \to -k} \frac{x+k}{(x+1)(x+2)...(x+n)}=\frac{1}{(-k+1)(-k+2)...(-k+n)}=\frac{(-1)^{(k-1)}}{(k-1)!(n-k)!}$$
so yes it's true that
$$\frac{1}{x(x+1)(x+2)...(x+n)}=\frac{1}{(n-1)!}\sum_{k=1}^n {n-1\choose k-1}\frac{(-1)^{(k-1)}}{x+k}$$

