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.