<s>[INST] Showing that $\frac 1 {(x+1)^2}=\frac 1 {(x+1)(x+2)}+\frac {1!} {(x+1)(x+2)(x+3)} + \frac {2!} {(x+1)(x+2)(x+3)(x+4)} + \cdots$ 
Prove that if $x>-1$ then
  $$\frac 1 {(x+1)^2}=\frac 1 {(x+1)(x+2)}+\frac {1!} {(x+1)(x+2)(x+3)} + \frac {2!} {(x+1)(x+2)(x+3)(x+4)} + \cdots$$

Could I have a hint for this? I tried writing the RHS as a power series and coming up with a differential equation but it doesn't really help as I end up with an intractable integral.
 [/INST] Hint: This is equivalent to 
$$\frac{1}{x} = \frac{0!}{(x+1)}+\frac{1!}{(x+1)(x+2)}+\frac{2!}{(x+1)(x+2)(x+3)}+\cdots.$$
Let $$f_n(x)=\frac{n!}{x(x+1)\cdots(x+n)}.$$
We have that
$$f_n(x)=f_{n-1}(x)-f_{n-1}(x+1),$$
and we wish to prove that
$$\frac{1}{x}=\sum_{n=0}^{\infty} f_n(x+1).$$
Can you take it from here?
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