TITLE: On the precise asymptotic scaling of $n!/(n-k)!$ as $n,k \to \infty$
QUESTION [7 upvotes]: On page 23 of [Erdős+Rényi 1960, "On the evolution of random graphs"], the following asymptotic formula is stated without proof:
$$ \binom{n}{k} \sim \frac{n^k \mathrm e^{-\frac{k^2}{2n} - \frac{k^3}{6n^2}}}{k!} $$
valid for $k \in o(n^{\text{[some illegible fraction]}})$. That is, we have 
$$ \frac{n!}{(n-k)!} \;=\; n^k \exp\left(-\tfrac{k^2}{2n} - \tfrac{k^3}{6n^2}\right) \cdot \Bigl[1 \pm o(1)\Bigr].$$
I was curious about what the limitations of the illegible upper bound on $k$ were for this approximation, and hoped that I could use at least some useful scaling for $k \in \Theta(n^{2/3})$, so I tried to rederive it. However, using Stirling's approximation for the factorial,
$$ n! = n^{n+\frac{1}{2}} \mathrm e^{-n}\cdot \Bigl[\sqrt{2\pi} + o(1)\Bigr], $$
the best that I could rederive was the following:
$$\begin{align}
\frac{n!}{(n-k)!}
\;&=\;
\frac{n^{n-k+\frac{1}{2}} n^k \mathrm e^{n-k}}{(n-k)^{n - k + \frac{1}{2}} \mathrm e^n} \cdot \Bigl[1 \pm o(1)\Bigr]
\\&=\;
n^k \left(1 - \frac{k}{n}\right)^{-n+k-\frac{1}{2}} \mathrm{e}^{-k} \cdot \Bigl[1 \pm o(1)\Bigr]\end{align}$$
which looks as though it should scale like
$$ \frac{n!}{(n-k)!} \;\;\stackrel{\;?\,}\sim\;\; n^k \exp\Bigl( - \tfrac{k^2}{n} + \tfrac{k}{2n} \Bigr),$$ for all $k \in o(n)$. This is close, but no cigar. Is there anything that I'm missing?

REPLY [3 votes]: As you say, if $n$ approaches infinity so that $k/n\to 0$, by Stirling's approximation
$$
k! n^{-k} \binom{n}{k} = (1 + o(1)) (1-\frac{k}{n})^{-(n-k)} e^{-k}. \qquad\ \ \  (*)
$$
Then, taking logarithms and expanding in a Taylor series with remainder gives
\begin{eqnarray*}
\log\left((1-\frac kn)^{-(n-k)} e^{-k}\right)&=&-k-(n-k)\log(1-\frac kn)\\
&=& -k + (n-k) \left(\sum_{1\le i\le j} \frac 1i (\frac kn)^i + O((\frac kn)^{j+1})\right)\\
&=& -\sum_{1\le i\le j-1} \frac 1 {i(i+1)} \frac{k^{i+1}}{n^i}+ O(\frac{k^{j+1}}{n^j})
\end{eqnarray*}
for any fixed $j\ge 1$.  Exponentiating and substituting this back into $(*)$ gives, as $n\to\infty$,
$$
\binom{n}{k} = (1 + o(1)) \frac{n^k}{k!} \exp\left(-\sum_{1\le i\le j-1}\frac{1}{i(i+1)} \frac{k^{i+1}}{n^i} \right),\ \ \ \text{where } k=o(n^{j/(j+1)}).
$$
The formula in the Erdős-Rényi paper is obtained by setting $j:=3$.  The illegible exponent should then be $3/4$.