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.
It should be noted one may also rewrite the summation and appeal directly to the Negative Binomial Series 7.4.1 to also prove this result.