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