Curious identity involving the number of perfect matchings of the complete graph

Question

Can you prove (preferably combinatorially) the following identity for the total number of perfect matchings of the complete graph $K_{2n}$, where the edges in the matching are ordered, i.e., $\binom{2n}{2,2,\ldots ,2} = \frac{(2n)!}{2^{n}} = n!(2n-1)(2n-3)\cdots 1$:

$$ \binom{2n}{2,2,\ldots ,2} = \prod_{k=1}^{n} \left[\binom{n+1}{2}-\binom{k}{2}\right] = n\prod_{j=1}^{n-1} [n+(n-1)+ \ldots + (n-j)]. $$

It is also immediate to show that the total number of matchings is $\prod_{i=2}^{n} [i(i+(i-1))]$.

Thus it suffices to show that for every $n\geq 1$, \begin{equation} n[n+(n-1)]\cdots [n+(n-1)+\cdots + 1] = [n(n+(n-1))][(n-1)((n-1)+(n-2))]\cdots [2(2+1)]. \end{equation}

Algebraic proof would be also of some help. I tried by induction unsuccessfully.

Answer 1

It is somewhat unclear from the question which $=$ indicate known identities and which indicate conjectured identities. I assume from the "it suffices" that what you want is a proof of $$n[n+(n-1)]\cdots [n+(n-1)+\cdots + 1] \stackrel?= [n(n+(n-1))][(n-1)((n-1)+(n-2))]\cdots [2(2+1)]$$ or, more succinctly, $$\prod_{k=1}^n \left(kn - \frac{k(k-1)}{2}\right) \stackrel?= n! (2n-1)!!$$

If we multiply both sides by $2^n$ we get $$\prod_{k=1}^n \left(2kn - k(k-1)\right) \stackrel?= (2n)!$$

But then $$\prod_{k=1}^n \left(2kn - k(k-1)\right) = \left(\prod_{k=1}^n k \right) \prod_{k=1}^n \left(2n - (k-1)\right) = \left(\prod_{k=1}^n k \right) \prod_{j=n+1}^{2n} j = (2n)!$$ with the substitution $j = 2n - (k-1)$.

Note as a corollary which I, at least, find interesting, that $$\prod_{k=1}^n \left(n - \frac{k-1}{2}\right) = (2n-1)!!$$

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This suggests that a combinatorial proof will have to treat the terms differently according to the parity of $k$:

$$\prod_{k=1}^n \left(kn - \frac{k(k-1)}{2}\right) = \underbrace{\left(\prod_{j=1}^{n/2} \color{red}{j} \color{blue}{(2n - 2j + 1)}\right)}_{k \textrm{ even}}\, \underbrace{\prod_{j=0}^{(n-1)/2} \color{blue}{(2j+1)} \color{\red}{(n - j)}}_{k \textrm{ odd}} = \color{red}{n!} \color{blue}{(2n-1)!!}$$