How do you go from the binomial to the Poisson?

In the limit that $p\rightarrow 0$ and $n»\mu$ (assume $n\rightarrow \infty$), the binomial becomes a Poisson (this one is relatively easy to prove). The binomial distribution can be written $B(x; n,p) = \frac{1}{x!}\frac{n!}{(n-x)!} p^x (1-p)^{-x}(1-p)^n$. The quantity $\frac{n!}{(n-x)!}=n(n-1)(n-2)...(n-x+2)(n-x+1)$ has $x$ factors, all close to $n$ for $n»x$: so this can be approximated $n^x$. Putting that together with $p^x$, you get $(np)^x=\mu^x$. For $p$ very small, you have $(1-p)^{-x}\sim 1+px\sim 1$. The rightmost term $(1-p)^n$, can be written $\lim_{p\rightarrow 0} [(1-p)^{1/p}]^\mu$, since $n=\mu/p$. Using the that $e^x = \lim_{\alpha\rightarrow \infty}(1+\frac{x}{\alpha})^\alpha = \lim_{1/\beta\rightarrow 0} (1+\beta x)^{1/\beta}$, we have that $\lim_{p\rightarrow 0} [(1-p)^{1/p}]^\mu=e^{-\mu}$. So overall, we have $\mathcal{P}(x;\mu) = \frac{\mu^x}{x!}e^{-\mu}$ in this limit.