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