Proof
\begin{equation*}
M(0) = \frac{1}{ \left ( 1-\mu 0 \right )^{r}} \frac{1}{1} = 1.
\end{equation*}
Continuing,
\begin{equation*}
M'(t) = \frac{r \mu}{ \left ( 1-\mu t \right )^{r+1}}
\end{equation*}
and therefore
\begin{equation*}
M'(0) = \frac{r \mu}{ \left ( 1-\mu 0 \right )^{r+1}} = r \mu.
\end{equation*}
Continuing with the second derivative,
\begin{equation*}
M''(t) = \frac{r(r+1) \mu^2}{ \left ( 1-\mu t \right )^{r+2}}
\end{equation*}
and therefore
\begin{equation*}
M''(0) = \frac{r(r+1) \mu^2}{ \left ( 1-\mu 0 \right )^{r+2}} = r(r+1) \mu^2 = r \mu^2 + r^2 \mu^2
\end{equation*}
which is the squared mean plus the variance for the poisson distribution.