Semifactorial Identity I was wondering if anyone had any insight on how to prove the following identity:

For all $m \in \mathbb{N}$ $$ \frac{1}{2m-1} + \frac{2m-2}{(2m-1)(2m-3)} + \frac{(2m-2)(2m-4)}{(2m-1)(2m-3)(2m-5)} + \cdots + \frac{(2m-2)!!}{(2m-1)!!} = 1$$

I attempted to rewrite and simplify the left hand side as:
$$ \sum_{k = 1}^m \frac{\frac{(2m-2)!!}{(2m-2k)!!}}{\frac{(2m-1)!!}{(2m-2k-1)!!}} = \sum_{k = 1}^m \frac{(2m-2)!! (2m-2k-1)!!}{(2m-2k)!! (2m-1)!!} = \frac{\left[2^{m-1} (m-1)! \right]^2}{(2m-1)!} \sum_{k=1}^m \frac{(2m-2k-1)!!}{(2m-2k)!!} $$
But I do not seem to be getting anywhere. Does anyone see how to prove the statement?

I believe this can be proven combinatorially by recasting it as
$$(2m-3)!! + (2m-2)(2m-5)!! + (2m-2)(2m-4)(2m-7)!! + \cdots + (2m-2)!! = (2m-1)!!$$
To do this, we note that $(2m-1)!!$ counts the number of perfect matchings of the complete graph $K_{2m}$. Label the vertices of $K_{2m}$ with the integers $1 \ldots 2m$, and consider any perfect matching of $K_{2m}$. Let $n$ be the integer such that $\{1,n\}$ is an edge of the matching, and color blue any edge of the matching which contains a vertex labelled less than $n$. Every perfect matching can be colored uniquely in this way, therefore the perfect matchings of $K_{2m}$ are the disjoint union of those matchings with $k$ blue edges, as $k$ varies from 1 to $m$ (note that the edge $\{1,n\}$ is always colored blue).
Let's count the number of matchings with $k$ blue edges. If $k=1$, then the blue edge must be $\{1,2\}$ since if $\{1,2\}$ is not an edge of the matching, both edges containing 1 and 2 would be blue. The remainder of the matching can be chosen in $(2m-3)!!$ ways, yielding $(2m-3)!!$ perfect matchings with one blue edge.
If $k=2$, then $\{1,2\}$ is not an edge of the matching, so the edge of the matching containing 2 must be colored blue. This edge can be picked in $2m-2$ ways, since the other vertex of the edge may be any other than 1 or 2. Since $k=2$, there can be only one more blue edge, which necessarily is an edge containing the vertex 1. This will be either $\{1,4\}$, if $\{2,3\}$ is the other blue edge, or $\{1,3\}$ otherwise. The remainder of the match can be chosen in $(2m-5)!!$ ways, yielding $(2m-2)(2m-5)!!$ perfect matchings with 2 blue edges.
For general $k$, we proceed as above. For each $i \in \{1\ldots k-1\}$, we create a blue edge by picking the smallest unmatched vertex greater than 1, and finding it a mate distinct from 1 that has not yet been matched; at each step, this can be done in $2m-2i$ ways ($2i-2$ already matched vertices, 1, and the vertex in question are excluded). After this process is complete, the $k^{th}$ edge must be created by connecting the vertex 1 to the remaining unmatched vertex with the smallest label, for which there is only one choice. The remainder of the perfect match can be made in $(2(m-k)-1)!!$ ways.
Adding up all of these counts as $k$ varies from 1 to $m$ yields the left-hand side of the identity above.
