Poisson as limit of Binomial distribution Suppose that we have a large number $n$ of independent trials, but the probability $p$ of success is very small, in such a way the the expectation $\mu=np$ of the number of successes is moderate. Keep $\mu=np$ fixed and let $n$ tend to infinity. Then the probability $r$ successes is:
$b(n,p;r)=b(n,\frac{\mu}{n};r)=C^n_r\times (\frac{\mu}{n})^r(1-\frac{\mu}{n})^{n-r}=\frac{\mu^r}{r!}\frac{n(n-1...(n-r+1)}{(n-\mu)^r}(1-\frac{\mu}{n})^n$
As $n\to\infty$ 
$b(n,p;r)\to\mu^r \frac{e^{-\mu}}{r!}$
After taking a course on measure theory I am studying probability theory as the application of measure theory but I am having some problems on the convergence of functions. I tried to use Taylor expansion to approximate the Binomial of a Poisson but unsuccessfully.
(1) How do I to get from $\frac{\mu^r}{r!}\frac{n(n-1...(n-r+1)}{(n-\mu)^r}(1-\frac{\mu}{n})^n$ to $\mu^r \frac{e^{-\mu}}{r!}$
(2)What is the difference between a Binomial and Poission distribution? Is it the $p$?
Thanks in advance!
 A: The Poisson Distribution may be regarded as a limiting form of the binomial distribution when number of trials $n\to\infty$, and probability of success in trial $p\to0$ but $np(=\mu)$ remains finite.
$
\lim\limits_{n\to\infty}\displaystyle\binom{n}{r}\left(\dfrac{\mu}{n}\right)^r\left(1-\dfrac{\mu}{n}\right)^{n-r}\\
=\lim\limits_{n\to\infty}\dfrac{n(n-1)(n-2)\cdots(n-r+1)}{r!}\left(\dfrac{\mu}{n}\right)^r\left(1-\dfrac{\mu}{n}\right)^{n-r}\\
=\lim\limits_{n\to\infty}\dfrac{1\left(1-\dfrac1n\right)\left(1-\dfrac2n\right)\cdots\left(1-\dfrac{r-1}{n}\right)}{r!}(\mu)^r\left(1-\dfrac{\mu}{n}\right)^{n-r}\\
=\dfrac{\mu^r}{r!}\cdot\lim\limits_{n\to\infty}\left[1\left(1-\dfrac1n\right)\left(1-\dfrac2n\right)\cdots\left(1-\dfrac{r-1}{n}\right)\right]\cdot\lim\limits_{n\to\infty}\left(1-\dfrac{\mu}{n}\right)^n\cdot\dfrac{1}{\lim\limits_{n\to\infty}\left(1-\dfrac{\mu}{n}\right)^r}\\
=\lim\limits_{n\to\infty}\left(1-\dfrac{\mu}{n}\right)^n\cdot\dfrac{\mu^r}{r!}\cdot\lim\limits_{n\to\infty}\left[1\left(1-\dfrac1n\right)\left(1-\dfrac2n\right)\cdots\left(1-\dfrac{r-1}{n}\right)\right]\cdot\dfrac{1}{\lim\limits_{n\to\infty}\left(1-\dfrac{\mu}{n}\right)^r}\\
=e^{-\mu}\cdot\dfrac{\mu^r}{r!}\cdot1\cdot\dfrac{1}{1}\\
=\dfrac{e^{-\mu}\mu^r}{r!}.
$
The practical utility of the result is that, when $n$ is sufficiently large and $p$ is quite small, but $np(=\mu)$ has a moderate value, the binomial probabilities $\displaystyle\binom{n}{r}\left(\dfrac{\mu}{n}\right)^r\left(1-\dfrac{\mu}{n}\right)^{n-r}$ can be well approximated by the corresponding Poisson Probabilities $\dfrac{e^{-\mu}\mu^r}{r!}$, which is much easier to compute.
A: You have
$$\frac{\mu^r}{r!}\frac{n(n-1)\cdots(n-r+1)}{(n-\mu)^r}
\left(1-\frac{\mu}{n}\right)^n.$$
Consider $r$ and $\mu$ as fixed and let $n\to\infty$. Then
$(n-\mu)^r\sim n^r$ and $n(n-1)\cdots(n-r+1)\sim n^r$ also. In short
$$\lim_{n\to\infty}\frac{n(n-1)\cdots(n-r+1)}{(n-\mu)^r}=1.$$
Also
$$\lim_{n\to\infty}\left(1-\frac{\mu}{n}\right)^n=e^{-\mu};$$
this is a special case of
$$\lim_{n\to\infty}\left(1+\frac{x}{n}\right)^n=e^x.$$
Therefore
$$\lim_{n\to\infty}\frac{\mu^r}{r!}\frac{n(n-1)\cdots(n-r+1)}{(n-\mu)^r}
\left(1-\frac{\mu}{n}\right)^n=e^{-\mu}\frac{\mu^r}{r!}.$$
