I think it is worth mentioning that a Poisson($\lambda$) pmf is the limiting pmf of a Binomial($n$,$p_n$) with $p_n = \lambda / n$.
One rather lengthy development can be found on this blog.
But, we can prove this economically here as well. If $X_n \sim \mathrm{Binomial}(n,\lambda/n)$ then for fixed $k$
$$
\begin{align}
\mathbb P(X_n = k) &= \frac{n!}{k!(n-k)!} \left(\frac{\lambda}{n}\right)^k \left(1-\frac{\lambda}{n}\right)^{n-k} \\ &= \underbrace{\frac{n! n^{-k}}{(n-k)!}}_{\to 1} \frac{\lambda^k}{k!}\underbrace{(1-\lambda/n)^n}_{\to e^{-\lambda}} \cdot \underbrace{(1-\lambda/n)^{-k}}_{\to 1} \>.
\end{align}
$$
The first and last terms are easily seen to converge to 1 as $n \to \infty$ (recalling that $k$ is fixed). So,
$$
\mathbb P(X_n = k) \to \frac{e^{-\lambda} \lambda^k}{k!} \,,
$$
as $n \to \infty$ since $(1-\lambda/n)^n \to e^{-\lambda}$.
In addition one has the normal approximation to the Binomial, i.e., Binomial($n$,$p$) $\approxeq^d \mathcal N(np, np(1-p))$. The approximation improves as $n \rightarrow \infty$ and $p$ stays away from 0 and 1. Obviously for the Poisson regime this is not the case (since there $p_n = \lambda / n \rightarrow 0$) but the larger $\lambda$ is the larger $n$ can be and still have a reasonable normal approximation.
