Proof
\begin{equation*}
M(0) = \left ( p e^0 + (1-p) \right )^n = 1^n = 1.
\end{equation*}
Taking the derivative with respect to t,
\begin{equation*}
M'(t) = n \left ( p e^t + (1-p) \right )^{n-1} p e^t
\end{equation*}
and evaluating at t=0 gives
\begin{equation*}
M'(0) = n \left ( p + (1-p) \right )^{n-1} p = n 1^{n-1} p = np.
\end{equation*}
Again, taking another derivative with respect to t,
\begin{equation*}
M''(t) = n(n-1) \left ( p e^t + (1-p) \right )^{n-2} p^2 e^{2t} + n \left ( p e^t + (1-p) \right )^{n-1} p e^t
\end{equation*}
and evaluating at t=0 gives
\begin{align*}
M''(0) & = n(n-1) ( p + (1-p))^{n-2} p^2 + n ( p + (1-p) )^{n-1} p \\
& = n(n-1)p^2 + np = (np)^2 + np - np^2 = np(1-p) + (np)^2.
\end{align*}