poisson distribution

Let $\lambda =np$ … expected number of successes (mean of the distribution).
If $X\sim$ binomial$(n,p)$ , then

$$\begin{array}{rl}P(X=k)& =\left(\genfrac{}{}{0ex}{}{n}{k}\right)p^k(1-p{)}^{n-k}\\ & =\frac{n!}{k!(n-k)!}{\left(\frac{\lambda }{n}\right)}^k{\left(1-\frac{\lambda }{n}\right)}^{n-k}\\ & =\frac{{\lambda }^k}{k!}\frac{n(n-1)(n-2)\cdots (n-k+1)}{n^k}{\left(1-\frac{\lambda }{n}\right)}^n{\left(1-\frac{\lambda }{n}\right)}^{-k}\\ \text{as }n\to \infty :& \\ & =\frac{{\lambda }^k}{k!}\text{cancelto}1\frac{n(n-1)(n-2)\cdots (n-k+1)}{n^k}{\left(1-\frac{\lambda }{n}\right)}^n\text{cancelto}1{\left(1-\frac{\lambda }{n}\right)}^{-k}\\ & =\frac{{\lambda }^k}{k!}{e}^{-\lambda }\end{array}$$