I am trying to follow a derivation of the moment generating function of the negative binomial distribution with parameters $k$ and $n$, but have found myself unable understand the following equivalence:
Clearly this implies that $$ \sum_{x=k}^\infty {{x-1}\choose{k-1}} (qe^t)^{x-k} = (1-qe^t)^{-k} $$ but I am unsure as to why this is the case
Replacing $x$ by $k+n$ gives $$ \sum_{n=0}^{\infty} \binom{k+n-1}{k-1} (qe^t)^n , $$ which looks quite binomial already. You then want to move the dependence on $k$ to the lower argument of the binomial coefficient, which can be done as follows: $$ \begin{align} \binom{k+n-1}{k-1} &= \frac{(k+n-1)!}{n!(k-1)!} \\ &= \frac{(k+n-1)(k+n-2) \dotsm (k+1)k }{n!} \\ &= (-1)^n \frac{(-k)(-k-1) \dotsm (k-n+1) }{n!} \\ &= (-1)^n \binom{-k}{n} , \end{align} $$ and it should be smooth sailing from here.