Proof
\begin{equation*}
M(0) = \frac{(p)^r}{(1 - (1-p))^r} = \frac{p^r}{p^r} = 1.
\end{equation*}
Taking the derivative with respect to t,
\begin{equation*}
M'(t) = -\frac{{\left(p e^{t} - e^{t} + 1\right)}^{r - 1} {\left(p - 1\right)} p^{r} r e^{t}}{{\left(p e^{t} - e^{t} + 1\right)}^{2 r}}
\end{equation*}
and evaluating at t=0 gives
\begin{equation*}
M'(0) = -\frac{{p}^{r - 1} {p - 1} p^{r} r }{{p}^{2 r}} = \frac{r(1-p)}{p}.
\end{equation*}
OOPS...don’t need the (1-p) on top. Again, taking another derivative with respect to t,
\begin{equation*}
M''(t) = -\frac{{\left(p e^{t} - e^{t}\right)} {\left(p e^{t} - e^{t} + 1\right)}^{r - 2} {\left(p - 1\right)} p^{r} {\left(r - 1\right)} r e^{t}}{{\left(p e^{t} - e^{t} + 1\right)}^{2r}}
\\ + \frac{2 {\left(p e^{t} - e^{t}\right)} {\left(p e^{t} - e^{t} + 1\right)}^{2r - 2} {\left(p - 1\right)} p^{r} r^{2} e^{t}}{{\left(p e^{t} - e^{t} + 1\right)}^{3r}}
\\ - \frac{{\left(p e^{t} - e^{t} + 1\right)}^{r - 1} {\left(p - 1\right)} p^{r} r e^{t}}{{\left(p e^{t} - e^{t} + 1\right)}^{2r}}
\end{equation*}
and evaluating at t=0 gives
\begin{align*}
M''(0) & = \frac{{\left(p^{2 \, r - 1} r - p^{2 \, r - 2} r - p^{2 \, r - 2}\right)} {\left(p - 1\right)} r}{p^{2 \, r}} \\
& = \left ( \frac{r(1-p)}{p^2} \right )^2 + \left ( \frac{r}{p} \right )^2.
\end{align*}