I'm trying to prove that, for $N\gg n$,
$P(N,n)=\frac{N!}{(N-n)!}\approx N^{n}$
I've tried two approaches,
1
$\frac{N!}{(N-n)!}=\frac{N\left(N-1\right)\cdots\left(N-n+1\right)\left(N-n\right)!}{\left(N-n\right)!}=N(N-1)\cdots(N-n+1)=\prod_{i=0}^{n-1}\left(N-i\right)$
I now understand that I have $n$ factors involving $N$ minus something which is to be small compared with $N$ so that I could say it approximates $N^{n}$ . But I should be able to state it formally.
2
$\frac{N!}{(N-n)!}=\exp\left(\ln\frac{N!}{(N-n)!}\right)=\exp\left(\ln N!-\ln\left(N-n\right)!\right)$
Doing the stirling approximation I get,
$\approx\exp\left(N\ln N-N-\left(N-n\right)\ln\left(N-n\right)+N-n\right)=N^{N}(N-n)^{n-N}e^{-n}$
Use Stirling approximation like this: $N!=N^Ne^{-N}$. Then you get:
$\frac{N!}{(N-n)!} \approx \frac{e^{-n}N^N}{(N-n)^{N-n}}=N^n e^{-n}\left(\frac{N}{N-n}\right)^{N-n}$
Then you can see that $\left(\frac{N}{N-n}\right)^{N-n}\approx e^n$ which solves your problem.