Simplifying a Product of Summations I have, for a fixed and positive even integer $n$, the following product of summations:
$\left ( \sum_{i = n-1}^{n-1}i \right )\cdot \left (  \sum_{i = n-3}^{n-1} i \right )\cdot \left ( \sum_{i = n-5}^{n-1}i \right )\cdot ... \cdot \left (\sum_{i=5}^{n-1}i \right )\cdot \left (\sum_{i=3}^{n-1}i \right )\cdot \left (\sum_{i=1}^{n-1}i \right )$
Where there are $\frac{n}{2}$ groups of summations multiplied together.
For example, consider the case where $n=4$ :
$\left ( \sum_{i = 3}^{3}i \right )\cdot \left (  \sum_{i = 1}^{3} i \right ) = \left ( 3 \right )\left ( 1+2+3 \right ) = 18$
I have tried in vain to simplify the product. Perhaps there are identities I could make use of.
Edit : I can expand the product to clarify:
$$\left ( n-1 \right )\cdot \left [ (n-3)+(n-2)+(n-1) \right ]\cdot \left [ (n-5)+...+(n-1) \right ]\cdot ... \cdot\left [3+4+...+(n-1)\right ]\cdot \left [1+2+...+(n-1) \right ]$$
From where I can see a $(n-1)^{\frac{n}{2}}$ term, but the others are quite jumbled.

Let $n=2m$. Your expression is then
$$\begin{align*}
&(2m-1)\cdot\frac32(2n-4)\cdot\frac52(2n-6)\cdot\ldots\cdot\frac{n-1}2n\\\\
&\qquad=1\cdot(2m-1)\cdot3(2m-2)\cdot5(2m-3)\cdot\ldots\cdot(2m-1)m\\\\
&\qquad=\prod_{k=1}^m(2k-1)(2m-k)\\\\
&\qquad=(2m-1)!!\frac{(2m-1)!}{(m-1)!}\\\\
&\qquad=\frac{(2m)!}{2^mm!}\cdot\frac{(2m-1)!}{(m-1)!}\\\\
&\qquad=\frac{(2m)!}{2^m}\binom{2m-1}m\\\\
&\qquad=\frac{n!}{2^{n/2}}\binom{n-1}{n/2}\;.
\end{align*}$$
