The Poisson distribution is actually a limiting case of a Binomial distribution when the number of trials, n, gets very large and p, the probability of success, is small. As a rule of thumb, if $n \ge 100$ and $np \le 10$, the Poisson distribution (taking $\lambda = np$) can provide a very good approximation to the binomial distribution.

This is particularly useful as calculating the combinations inherent in the probability formula associated with the binomial distribution can become difficult when $n$ is large.

To better see the connection between these two distributions, consider the binomial probability of seeing $x$ successes in $n$ trials, with the aforementioned probability of success, $p$, as shown below.

$$P(x)={}_nC_x p^x q^{n-x}$$Let us denote the expected value of the binomial distribution, $np$, by $\lambda$. Note, this means that

$$p=\frac{\lambda}{n}$$and since $q=1-p$,

$$q=1-\frac{\lambda}{n}$$Now, if we use this to rewrite $P(x)$ in terms of $\lambda$, $n$, and $x$, we obtain

$$P(x) = {}_nC_x \left( \frac{\lambda}{n} \right)^x \left( 1-\frac{\lambda}{n} \right)^{n-x}$$Using the standard formula for the combinations of $n$ things taken $x$ at a time and some simple properties of exponents, we can further expand things to

$$P(x) = \frac{n(n-1)(n-2) \cdots (n-x+1)}{x!} \cdot \frac{\lambda^x}{n^x} \left( 1 - \frac{\lambda}{n} \right)^{n-x}$$Notice that there are exactly $x$ factors in the numerator of the first fraction. Let us swap denominators between the first and second fractions, splitting the $n^x$ across all of the factors of the first fraction's numerator.

$$P(x) = \frac{n}{n} \cdot \frac{n-1}{n} \cdots \frac{n-x+1}{n} \cdot \frac{\lambda^x}{x!}\left( 1 - \frac{\lambda}{n} \right)^{n-x}$$Finally, let us split the last factor into two pieces, noting (for those familiar with Calculus) that one has a limit of $e^{-\lambda}$.

$$P(x) = \frac{n}{n} \cdot \frac{n-1}{n} \cdots \frac{n-x+1}{n} \cdot \frac{\lambda^x}{x!}\left( 1 - \frac{\lambda}{n} \right)^n \left( 1 - \frac{\lambda}{n} \right)^{-x}$$It should now be relatively easy to see that if we took the limit as $n$ approaches infinity, keeping $x$ and $\lambda$ fixed, the first $x$ fractions in this expression would tend towards 1, as would the last factor in the expression. The second to last factor, as was mentioned before, tends towards $e^{-\lambda}$, and the remaining factor stays unchanged as it does not depend on $n$. As such, $$\lim_{n \rightarrow \infty} P(x) = \frac{e^{-\lambda} \lambda^x}{x!}$$

Which is what we wished to show.