# Proof of inverse factorial decomposition $\frac{1}{n!}=\sum\limits^n_{i=1}\frac{(-1)^{n+i}}{(i-1)!(n+1-i)!}$

Doing research in probability modelling I obtained a decomposition of inverse factorial:

$$\frac{1}{n!}=\sum\limits^n_{i=1}\frac{(-1)^{n+i}}{(i-1)!(n+1-i)!}$$

How can it be proven directly? In what literature can I find it?

P.S. For odd $$n$$ the equality is trivial because all terms except $$\frac{1}{n!}$$ vanish. The trick is to prove it for even $$n$$.

Hint: Note that the formula is equivalent to $$1=\sum\limits_{i=1}^{n}(-1)^{n+i}\cdot \frac{n!}{(i-1)!(n+1-i)!},$$ i.e. $$1=(-1)^n\sum\limits_{i=1}^{n}(-1)^{i}\binom{n}{i-1}.$$
To show this, try using the fact that $$\sum\limits_{k=0}^{n}(-1)^k \binom{n}{k}=0$$ (which you can show using the binomial theorem).