Furthermore, to obtain an even power of x will require an even number of odd powers and an odd power of x will require an odd number of odd powers. So, the coefficient of the odd terms stays odd and the coefficient of the even terms remains even. Therefore,
\begin{equation*}
\left ( \frac{1}{1+x} \right )^n = \sum_{k=0}^{\infty} {(-1)^k \binom{n + k - 1}{k} x^k}
\end{equation*}
Similarly,
\begin{equation*}
\left ( \frac{1}{1-x} \right )^n = \left ( \sum_{k=0}^{\infty} {x^k} \right )^n = \sum_{k=0}^{\infty} {\binom{n + k - 1}{k} x^k}
\end{equation*}