The lifetime (in years) of a bulb is distributed as an $Exp(1)$ random variable. Using Poisson approximation to Binomial distribution, the probability that out of 50 randomly chosen bulbs, at most 1 fails within one month is ____?
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Well, this is a basic fact of the exponential function $e^x$.
One definition of $e$ is the limit $\lim_{n\to\infty}(1+\frac1n)^n$. By a monotonicity argument one can prove $\lim_{x\to\infty}(1+\frac1x)^x=e$ where $x$ now ranges the real numbers.
Also note that $1-\frac1x=\frac{x-1}x=1/\frac x{x-1}=1/(1+\frac1y)=(1+\frac1y)^{-1}$ where $y=x-1$.
So, one has the following:
$$\begin{aligned}
\lim_{x\to\infty}(1-\frac1x )^x &= \lim_{y\to\infty}(1+\frac1y )^{-(y+1)}
\\
&=\lim_{y\to\infty}(1+\frac1y)^{-y}\times\lim_{y\to\infty}(1+\frac1y)^{-1}
\\
&=e^{-1}\times1=e^{-1}\,.
\end{aligned} $$
From here, assuming $\lambda>0$, $$\begin{aligned} e^{-\lambda}=(e^{-1})^\lambda &= \lim_{x\to\infty}(1-\frac1x)^{\lambda x} &\to{\ z:=\lambda x} \\ &= \lim_{z\to\infty}(1-\frac\lambda z)^z\,. \end{aligned} $$
In consequence, we have this limit for every sequence $z_n\to\infty$ written in place of $z$ and limiting on the natural $n\to\infty$. In particular, this also holds for $z_n=n$.
Note that we had to take the turnaround for arbitrary real numbers instead of integers only because of the exponent $\lambda$.
The short answer is that the Poisson approximation is faster and easier to compute and reason about, and among other things tells you approximately how big the exact answer is.
Here's a simple example: suppose you're trying to get something to happen in a video game that is rare; maybe it happens 1% of the time you do something, independently. You'd like to know how likely it is to happen at least once if you try, say, 100 times. Here we have $p = \frac{1}{100}, n = 100$ and so the binomial distribution gives us an exact answer, namely
$$1 - \left( 1 - \frac{1}{100} \right)^{100}.$$
But how big is this? Do you know off the top of your head? Is it, say, bigger or less than 50%?
The Poisson approximation answers this question quickly and easily: in this special case, it amounts to the approximation
$$\left( 1 - \frac{1}{100} \right)^{100} \approx e^{-1} \approx 0.368 \dots $$
which gives
$$1 - \left( 1 - \frac{1}{100} \right)^{100} \approx 1 - e^{-1} \approx 0.632 \dots $$
so we get that the odds are about 63% that we'll succeed at least once, which is bigger than 50% but maybe smaller than you might hope.
We learn something else too: the Poisson approximation tells us more generally that the odds of success are approximately a function of the product $np = \lambda$ only (which is the expected number of successes), so that e.g. if we had $p = \frac{1}{1000}$ and $n = 1000$ the answer would still be about 63%. This is valuable information and not entirely obvious from the exact answer, and knowing it saves you from having to recompute a bunch of binomials.
Sometimes $n$ can get large enough that it would actually be infeasible to calculate the exact binomial answer. For example, suppose $n = 10^{25}, p = 10^{-25}$; numbers this big regularly appear in physics or chemistry since Avogadro's number is so large. I can confidently say that the answer is still about 63% even though it is no longer feasible to exactly calculate $(1 - p)^n$ (just try it!). The funny thing here is that the larger $n$ gets the harder it becomes to exactly compute the binomials, but the more accurate the Poisson approximation gets; for numbers this large it is for all intents and purposes basically exact.
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