Proof
\begin{equation*}
M(0) = \frac{1}{1-\mu 0} = 1.
\end{equation*}
Continuing,
\begin{equation*}
M'(t) = \frac{\mu}{ \left ( 1-\mu t \right )^2}
\end{equation*}
and therefore
\begin{equation*}
M'(0) = \frac{\mu}{ \left ( 1-\mu 0 \right )^2} = \mu.
\end{equation*}
Continuing with the second derivative,
\begin{equation*}
M''(t) = \frac{2 \mu^2}{ \left ( 1-\mu t \right )^3}
\end{equation*}
and therefore
\begin{equation*}
M''(0) = \frac{2 \mu^2}{ \left ( 1-\mu 0 \right )^3}= 2 \mu^2 = \mu^2 + \mu^2.
\end{equation*}
which is the squared mean plus the variance for the poisson distribution.