How can we prove mgf of sample proportion of binomial distribution converges to exp(pt)?

$S_{n}$ 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

$$lim_{n to +infty}[1+p(1+t/n + (t/n)^2/2! + … -1)]^n=
lim_{n to +infty}(1+pt/n)^n?$$