Stirling's approximation proof?

[Mathsboi]

I read this in a book (it was stats and about poisson approx to normal)
Given was this:

$$n(n-1)(n-2) \cdots (n-r+1) = \frac{n!}{(n-r)!} \approx n^r$$
Stating that "Stirling's approximation" had been used.
So I looked the up and found:

$$\ln n! \approx n\ln n - n\$$


In the poisson distribution n is very large and $$r$$ is very small compared to $$n$$ so all the terms in the given equation approximate to $$n$$... This gives me my $$\approx n^r$$

But I just wondered where the Stirling equation comes into it...

$$\ln (\frac{n!}{(n-r)!}) = \ln(n!) - \ln((n-r)!) \$$
$$\approx n\ln n - n - \left[ (n-r)\ln((n-r)) - (n-r) \right]\$$
$$\approx n\ln n - n - (n-r)\ln((n-r)) + n - r \$$
$$\approx n\ln n - (n-r)\ln((n-r)) -r \$$
...
That's as far as I got...

$$\approx \ln (n^n) - \ln((n-r)^{(r-n)}) -r \$$

Unless taking logs, instead of to base e, to base n...

$$\approx Log_n (n^n) - Log_n ((n-r)^{(r-n)}) -r \$$
Then...
$$Log_n (n^n) - Log_n ((n-r)^{(r-n)}) = r\$$
$$n^n - (n-r)^{(n-r)} = n^r\$$

^ not sure if that's correct though

Can anyone help?

 

[HallsofIvy]

This is "number theory", not "algebra and linear algebra" so I am moving it.

 

[hamster143]

$$\ln(n-r) \approx ln(n) - r/n$$

$$n\ln n - (n-r)\ln((n-r)) -r \approx n \ln n - n \ln n + n r/n + r (\ln n - r/n) - r\approx r \ln n - r^2/n$$