Where does this equation for $n\binom{\binom{n-1}{2}}{m}$ come from? I'm reading a proof and am having trouble seeing why the following two lines are true:
\begin{align*}
n\frac{\binom{\binom{n-1}{2}}{m} }{ \binom{\binom{n}{2}}{m}}
&= n \left(\frac{n-2}{n}\right)^m \prod_{i=1}^{m-1} \left(1 - \frac{4i}{n(n-1)(n-2)-2i(n-2)} \right) \\
&= n \left(\frac{n-2}{n}\right)^m \left( 1+ O\left(\frac{(\log n)^2}{n}\right) \right)
\end{align*}
I have tried breaking down the initial binomial term, but it gets messy very quickly. Can someone shed some light here?
Edit: I forgot to add what is probably a relevant fact: here $m=\frac{1}{2}n(\log n+w(n))$ and $w(n)=o(\log n)$.
 A: Here is the first part:

The following is valid
  \begin{align*}
\binom{N}{m}&=\frac{N(N-1)\cdots(N-m+1)}{m!}\\
&=\frac{N^m}{m!}\left(1-\frac{1}{N}\right)\cdots\left(1-\frac{m -1}{N}\right)\\
&=\frac{N^m}{m!}\prod_{i=1}^{m-1}\left(1-\frac{i}{N}\right)
\end{align*}

With $N=\binom{n-1}{2}$, resp. $\binom{n}{2}$ we obtain

\begin{align*}
&\binom{\binom{n-1}{2}}{m}\binom{\binom{n}{2}}{m}^{-1}\\
&\quad=\left(\frac{1}{m!}\binom{n-1}{2}^m\prod_{i=1}^{m-1}\left(1-\frac{i}{\binom{n-1}{2}}\right)\right)
\left(m!\binom{n}{2}^{-m}\prod_{i=1}^{m-1}\left(1-\frac{i}{\binom{n}{2}}\right)^{-1}\right)\tag{1}\\
&\quad=\left(\frac{n-2}{n}\right)^m\prod_{i=1}^{m-1}\left(1-\frac{2i}{(n-1)(n-2)}\right)
\left(1-\frac{2i}{n(n-1)}\right)^{-1}\\
&\quad=\left(\frac{n-2}{n}\right)^m\prod_{i=1}^{m-1}
\left(\frac{n(n-1)(n-2)-2ni}{n(n-1)(n-2)}\right)\left(\frac{n(n-1)(n-2)-2i(n-2)}{n(n-1)(n-2)}\right)^{-1}\\
&\quad=\left(\frac{n-2}{n}\right)^m\prod_{i=1}^{m-1}
\frac{n(n-1)(n-2)-2ni}{n(n-1)(n-2)-2i(n-2)}\\
&\quad=\left(\frac{n-2}{n}\right)^m\prod_{i=1}^{m-1}
\frac{n(n-1)(n-2)-2i(n-2)-4i}{n(n-1)(n-2)-2i(n-2)}\\
&\quad=\left(\frac{n-2}{n}\right)^m\prod_{i=1}^{m-1}
\left(1-\frac{4i}{n(n-1)(n-2)-2i(n-2)}\right)\\
\end{align*}

Comment:


*

*In (1) we use the identity from above for both binomial coefficients.

