binomial formula for negative integer powers

For negative integer powers, the binomial formula can be written in terms of binomial coefficients like so:

$${(1-x)}^{-n}=\sum _{m=1}^{\mathrm{\infty }}\left(\genfrac{}{}{0pt}{}{m+n-1}{n-1}\right)x^m$$

Proof:   We shall prove this by induction on $n$. First, note that, if $n=1$, then $\left(\genfrac{}{}{0pt}{}{m}{0}\right)=1$, so our formula reduces to

$${(1-x)}^{-1}=\sum _{m=1}^{\mathrm{\infty }}{x}^m,$$

which is the formula for the sum of an infinite geometric series.

Next, suppose that the formula is valid for a certain value of $n$. Then we have

$${(1-x)}^{-n-1}={(1-x)}^{-1}{(1-x)}^{-n}=\left(\sum _{k=0}^{\mathrm{\infty }}{x}^k\right)\left(\sum _{m=0}^{\mathrm{\infty }}\left(\genfrac{}{}{0pt}{}{m+n-1}{n-1}\right)x^m\right)$$

The product of sums can be rewritten as the following double sum:

$$\sum _{m=0}^{\mathrm{\infty }}\sum _{k=0}^m\left(\genfrac{}{}{0pt}{}{n+k-1}{n-1}\right)x^m$$

The easiest way to see this is by rearranging the double sum as follows and adding columns

To evaluate the finite sums, we shall use the following identity for binomial coefficients. (See the entry http://planetmath.org/node/273“binomial coefficient” for more information about this identity.)

$$\sum _{k=0}^m\left(\genfrac{}{}{0pt}{}{n+k-1}{n-1}\right)=\left(\genfrac{}{}{0pt}{}{m+n}{n}\right)$$

Inserting this result value for the finite sum back into the double sum, we obtain

$${(1-x)}^{-n-1}=\sum _{m=0}^{\mathrm{\infty }}\left(\genfrac{}{}{0pt}{}{m+n}{n}\right)x^m.$$

Q.E.D.