$X_n = \operatorname{Bin}(n,\frac{\lambda}{n})$ for $\lambda > 0$. Show $M_{X_n}(t)$ converges to the MGF of Poisson($\lambda)$ as $n\rightarrow \infty$
So far I got
MGF of $$\operatorname{Poi}(\lambda)=e^{-\lambda}\sum_{x=0}^\infty(e^tx)^x/x!$$
MGF of $$\operatorname{Bin}(n,\frac{\lambda}{n})=\sum_{x=0}^{\infty}e^{tx}{n\choose x}(\frac{\lambda}{n})^x(1-\frac{\lambda}{n})^{n-x}$$
$$\lim_{n\to\infty} \sum_{x=0}^{\infty}e^{tx}{n\choose x}(\frac{\lambda}{n})^x(1-\frac{\lambda}{n})^{n-x}$$
I can see that $n \choose x$ as n goes to infinity will be $\frac1{x!}$, but I don't know how they got the $e^{-\lambda}$ and $x^x$. I could be wrong but I feel like $(1-\frac{\lambda}{n})^{n-x}$ just becomes 1 and $(\frac{\lambda}{n})^x$ becomes 0, which wouldn't really work out.
The moment generating function of a Poisson distribution with parameter $\lambda$ is
$$M_P(t)=\sum_{n=0}^\infty e^{nt}\frac{e^{-\lambda}\lambda^n}{n!}=e^{-\lambda}\sum_{n=0}^\infty \frac{(e^{t}\lambda)^n}{n!}=e^{-\lambda}e^{e^t\lambda}=e^{(e^t-1)\lambda}$$
The moment generating function of a Binomial distribution with parameters $n$ and $p$ is
$$M_B(t)=\sum_{k=0}^n e^{kt}{n\choose k}p^k(1-p)^{n-k}=\sum_{k=0}^n {n\choose k}(e^tp)^k(1-p)^{n-k}=(e^tp+1-p)^n$$
Now let $p=\dfrac{\lambda}{n}$ in the latter, the MGF is
$$M_B(t)=\left(e^t\frac\lambda n+1-\frac\lambda n\right)^n=\left(1+\frac{(e^t-1)\lambda}{n}\right)^n$$
Which tends to $e^{(e^t-1)\lambda}$ as $n\to\infty$.