$S_{n}$ follows Binomial(n,p).
$X_{n}$ is the sample proportion which is $X_{n} = S_{n}/n$.
How can we prove $\lim_{n \to +\infty} M_n{(t)} = e^{pt}$ ?
What I found is
\begin{align} \lim_{n \to +\infty} M_n{(t)} &=\lim_{n \to +\infty}(pe^{t/n} + q)^n \\ &=\lim_{n \to +\infty}(pe^{t/n} + 1-p)^n\\ &=\lim_{n \to +\infty}[1 + p(e^{t/n}-1)^n]\\ &=\lim_{n \to +\infty}[1+p(1+t/n + (t/n)^2/2! + \cdots -1)]^n\\ &=\lim_{n \to +\infty}(1+pt/n)^n\\ &=e^{pt} \end{align}
But how can we know
$$\lim_{n \to +\infty}[1+p(1+t/n + (t/n)^2/2! + ... -1)]^n= \lim_{n \to +\infty}(1+pt/n)^n?$$
$\frac {e^{x}-1} x \to 1$ as $x \to 0$. Given $\epsilon >0$ there exist $\delta >0$ such that $x \leq (e^{x}-1) \leq (1+\epsilon) x$ if $0 \leq x <\delta$. This gives $(\frac t n) \leq (e^{t/n}-1)\leq (1+\epsilon) (\frac t n)$ for $n$ sufficiently large and $t \geq 0$. Can you finish the argument now?