\begin{align}M_x\left(t \right)&=\sum_{x=k}^\infty{\binom{x-1}{k-1}}p^kq^{x-k}e^{tx}\\& = (e^tp)^k\sum_{x=0}^\infty\binom{k-1+x}{k-1}(e^tq)^x\end{align}
How do we get $\sum_{x=0}^\infty\binom{k-1+x}{k-1}(e^tq)^x = (1+qe^t)^{-k} ?$
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A google search gives these links: https://math.stackexchange.com/questions/848449/deriving-moment-generating-function-of-the-negative-binomial, https://math.stackexchange.com/questions/2830021/mgf-of-the-negative-binomial-distribution,https://math.stackexchange.com/questions/2180501/moment-generating-function-negative-binomial-alternative-formula. – StubbornAtom Apr 11 '19 at 07:35
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@StubbornAtom I did that. It didn't help much. – Maadhav Apr 11 '19 at 07:36