Real Analysis: Differential Equations (Leibnitz Formula)

Let I be an open interval and n be a natural number. Suppose that both f:I->R and g:I->R have n derivatives. Prove that fg:I->R has n derivatives, and we have the following formula called Leibnitz's formula:

(fg)^n(x) = the sum as k=0,1,2,...n of(n choose k)f^k(x)g^(n-k)(x) for all x in I.

Write the formula out explicitly for n=2 and n=3.