For negative integer powers, the binomial formula can be written in terms of binomial coefficients like so: $$(1 - x)^{-n} = \sum_{m = 1}^\infty \binom{m+n-1}{n-1} x^m$$

Proof: We shall prove this by induction on $n$ . First, note that, if $n=1$ , then $\binom{m}{0} = 1$ , so our formula reduces to $$(1 - x)^{-1} = \sum_{m = 1}^\infty x^m ,$$ 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 $$(1 - x)^{-n-1} = (1 - x)^{-1} (1 - x)^{-n} = \left( \sum_{k = 0}^\infty x^k \right) \left( \sum_{m = 0}^\infty {m+n-1 \choose n-1} x^m \right)$$ The product of sums can be rewritten as the following double sum: $$\sum_{m = 0}^\infty \sum_{k = 0}^m {n+k-1 \choose n-1} x^m$$ The easiest way to see this is by rearranging the double sum as follows and adding columns $$\begin{matrix} x^0 \sum_{m = 0}^\infty \binom{m+n-1}{n-1} x^m = & \binom{n-1}{n-1} & + & \binom{n}{n-1} x & + & \binom{n+1}{n-1} x^2 & + & \binom{n+2}{n-1} x^3 & + & \binom{n+3}{n-1} x^4 & + & \cdots \\ x^1 \sum_{m = 0}^\infty \binom{m+n-1}{n-1} x^m = & & & \binom{n-1}{n-1} x & + & \binom{n}{n-1} x^2 & + & \binom{n+1}{n-1} x^3 & + & \binom{n+2}{n-1} x^4 & + & \cdots \\ x^2 \sum_{m = 0}^\infty \binom{m+n-1}{n-1} x^m = & & & & & \binom{n-1}{n-1} x^2 & + & \binom{n}{n-1} x^3 & + & \binom{n+1}{n-1} x^4 & + & \cdots \\ x^3 \sum_{m = 0}^\infty \binom{m+n-1}{n-1} x^m = & & & & & & & \binom{n-1}{n-1} x^3 & + & \binom{n}{n-1} x^4 & + & \cdots \\ . & . & . & . & . & . & . & . & . & . & . & . \end{matrix}$$ To evaluate the finite sums, we shall use the following identity for binomial coefficients. (See the entry ``binomial coefficient'' for more information about this identity.) $$\sum_{k = 0}^m \binom{n+k-1}{n-1} = \binom{m + n}{n}$$ Inserting this result value for the finite sum back into the double sum, we obtain $$(1 - x)^{-n-1} = \sum_{m = 0}^\infty \binom{m + n}{n} x^m.$$