In the book "Introduction to probability and statistics for engineers", I found one explanation for how Poisson random variable can be used to approximate the binomial distribution, it goes like this:
Suppose that X is a binomial random variable with parameters (n, p) and let λ = np, then
$$ \begin{align*} P\{X =i\} &= \frac{n!}{(n - 1)! i!} p^{i} (1 - p)^{n - i} \\&= \frac{n!}{(n - 1)! i!}(\frac{\lambda }{n})^{i}(1 - \frac{\lambda }{n})^{n - i} \\&= \frac{n(n-1)...(n-i+1)}{n^{i}}\frac{\lambda^{i} }{n!}\frac{(1 - \lambda/n)^{n}}{(1 - \lambda/n)^{i}} \end{align*} $$
Now, for n large and p small,
$$ \begin{align*} &(1 - \frac{\lambda }{n})^{n} \approx e^{\lambda} &&& \frac{n(n-1)...(n-i+1)}{n^{i}} \approx 1 &&&& (1 - \frac{\lambda }{n})^{i} \approx 1& \end{align*} $$
Then:
$$ P\{X = i\} \approx e^{-\lambda}\frac{\lambda^{i}}{i!} $$
What I don't understand is this approximation:
$$ \frac{n(n-1)...(n-i+1)}{n^{i}} \approx 1 $$
My understand is that if $i$ is also large enough, let's say $i \approx n$ then the LHS express will approximate to $0$ instead of $1$?
$$A=\frac{n(n-1)...(n-i+1)}{n^{i}} =\frac{\Gamma (n+1)}{n^i\,\Gamma (n+1-i)}$$
Take logarithms and use Stirling approximation for $n \gg i$ to obtain $$\log(A)=-\frac{(i-1) i}{2 n}-\frac{(i-1) i (2 i-1)}{12 n^2}+O\left(\frac{1}{n^3}\right)$$ So, $\log(A) \to 0^-$ and then $A \to 1^-$.