So, let’s replace \(\frac{a}{b} = x\) and ignore for a while the term factored out. Then, we only need to show
\begin{equation*}
\sum_{k=0}^{\infty} {(-1)^k \binom{n + k - 1}{k} x^k} = \left ( \frac{1}{1+x} \right )^n
\end{equation*}
However
\begin{align*}
\left ( \frac{1}{1+x} \right )^n & = \left ( \frac{1}{1 - (-x)} \right )^n \\
& = \left ( \sum_{k=0}^{\infty} {(-1)^k x^k} \right )^n
\end{align*}