I am trying to take the limit of the following fraction :
$$ \lim_{N \to\infty} \frac { N !}{(N-r)!} $$
Attempts : I tried using the Stirling approximation $\ln(n!) =n \ln n - n $ but I figured it doesn't work for $n\to\infty$ and neither does the more accurate $n! = \sqrt{2 \pi n} (\frac{n}{e})^n$ for the same reason.
Either way , in both cases I get $e^{-r} N^r$ . I should note here that $r \in \mathbb{Z}^+ $.
Thank you in advance.
For all $r > 0$, $$\dfrac{N!}{(N-r)!}=N(N-1)(N-2)\cdots(N-r+1) \to \infty\text{.}$$ If $r = 0$, $$\dfrac{N!}{(N-r)!} = \dfrac{N!}{N!} = 1$$ and for all $r < 0$, let $s = -r$. Then $(N-r)! = (N+s)! > N!$ and $$\dfrac{N!}{(N-r)!} = \dfrac{N!}{(N+s)!}=\dfrac{N(N-1)\cdots 1}{(N+s)(N+s-1)\cdots (N+1)N(N-1)\cdots 1} = \dfrac{1}{(N+s)(N+s-1)\cdots (N+1)} \to 0\text{.}$$