Using the
a previous theorem 7.4.5 justifying the Negative Binomial probability function with
\(a = p e^t\) and
\(b = 1-p\) and by changing variables to
\(u = x-r\) gives
\begin{align*}
M(t) & = \sum_{u=0}^{\infty} e^{tu} \binom{u+r - 1}{r-1}(1-p)^{u}p^r\\
& = p^r \sum_{u=0}^{\infty} \binom{u + r - 1}{r-1}(e^t(1-p))^u \\
& = \frac{ p^{r}}{(1 - e^t(1-p))^r} \sum_{u=0}^{\infty} \binom{u + r - 1}{r-1}(1 - e^t(1-p))^r (e^t(1-p))^u \\
& = \frac{p^{r}}{(1 - e^t(1-p))^r}
\end{align*}
noting that the last summation is the the sum of a negative binomial probability function over its entire range.