Physicist: Need help understanding a line in a derivation The derivation I'm looking at is the Poisson distribution in M. Fox's "Quantum Optics" Chapter 5.

Now by Stirling's Formula:
$\lim_{N \to \infty} [\ln N!] = N \ln N - N $
we can see that
$\displaystyle\lim_{N \to \infty} \big[\ln \big(\frac{N!}{N^n(N-n)!}\big)\big] = 0$

It's not clear to me why this limit equates to zero. Am I missing something obvious?
 A: $$\begin{align}\log{\left (\frac{N!}{N^n (N-n)!} \right )} &= \log{N!} - n \log{N} - \log{(N-n)!} \\ &\approx N \log{N}-N - n \log{N} - (N-n) \log{(N-n)} + (N-n) \text {[By Stirling's Formula]}\\ &= (N-n) \log{N} - (N-n) \log{(N-n)} - n \\ &= (N-n) \log{\left (\frac{N}{N-n} \right )} - n \\ &= -(N-n) \log{\left (1-\frac{n}{N} \right )} - n \\ &\approx -N \left ( -\frac{n}{N}\right ) - n\end{align}$$
Thus the limit is indeed $0$ as $N \to \infty$.  
A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
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\begin{align}
{N! \over N^{n}\pars{N - n}!} &
\,\,\,\stackrel{\mrm{as}\ N\ \to\ \infty}{\sim}\,\,\,
{\root{2\pi}N^{N + 1/2}\expo{-N} \over
N^{n}\bracks{\root{2\pi}\pars{N - n}^{N - n + 1/2}
\expo{-\pars{N - n}}}}
\\[5mm] & =
{N^{N + 1/2}\expo{-N} \over
N^{n}\bracks{N^{N - n + 1/2}\pars{1 - n/N}^{N - n + 1/2}\expo{-\pars{N - n}}}}
\\[5mm] & \stackrel{\mrm{as}\ N\ \to\ \infty}{\sim}\,\,\,
{\expo{-n} \over \pars{1 - n/N}^{N}}
\,\,\,\stackrel{\mrm{as}\ N\ \to\ \infty}{\sim}\,\,\,\bbx{\large 1}
\end{align}
