Nifty identity with a really cool proof

Hey, I came across this in a textbook on algebra. It was in the appendix on induction, that I thought that I would just gloss over. Do not, however, use induction, as there is a much more elegant way!(This really just illistrates how great a certain proof of a certain famous theorem is)

Let f, g be n-differenciable functions on the reals. Let (n, r) denote the binomial coefficent. n is a positive integer. $\displaystyle f^i$ denote the ith derivitive of f.

Prove that $\displaystyle \displaystyle \sum_{i=0}^{i=n}(n,i)f^ig^{n-i}=(fg)^{n}$