Past exam of the mathematics course of the University of Cambridge / 2017 / ia / Paper 2 / 9F / b / Solution

For fixed $k$ and sufficiently large $N$,The finite product and last factor tend to one, while $(1-\lambda /N{)}^N\to e^{-\lambda }$. Thus the Poisson limit theorem givesThese numbers form a probability mass function because they are nonnegative and their sum is $e^{-\lambda }{\sum }_{k\ge 0}{\lambda }^k/k!=1$, by the exponential series. They define the Poisson distribution of parameter $\lambda$. Only $N\ge \lambda$ is needed to make the binomial parameter valid, which is sufficient for the limit.