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?

$$\scriptsize\begin{align}
&\;\;\;\frac 1{2m-1}+\frac {2m-2}{(2m-1)(2m-3)}+\frac{(2m-2)(2m-4)}{(2m-1)(2m-3)(2m-5)}+\cdots+\frac {(2m-2)!!}{(2m-1)!!}\\\\
&=\frac 1{2m-1}\left(1+\frac {2m-2}{2m-3}\left(1+\frac {2m-4}{2m-5}\left(1+\;\;\cdots\;\;\ \left(1+\frac 65\left(1+\frac 43\left(1+\frac 21\right)\right)\right)\right)\cdots+\right)\right)\\\\
&=\frac 1{2m-1}\left(1+\frac {2m-2}{2m-3}\left(1+\frac {2m-4}{2m-5}\left(1+\;\;\cdots\;\;\ \left(1+\frac 65\left(1+\frac 43\cdot \color{blue}3\right)\right)\right)\cdots+\right)\right)\\\\
&=\frac 1{2m-1}\left(1+\frac {2m-2}{2m-3}\left(1+\frac {2m-4}{2m-5}\left(1+\;\;\cdots\;\;\ \left(1+\frac 65\cdot \color{blue}5\right)\right)\cdots+\right)\right)\\\\
&\qquad\qquad \vdots\qquad\qquad\qquad\qquad\qquad \;\unicode{x22F0} \\\\
&=\frac 1{2m-1}\left(1+\frac {2m-2}{2m-3}\left(1+\frac {2m-4}{2m-5}\cdot \color{blue}{(2m-5)}\right)\right)\\\\
&=\frac 1{2m-1}\left(1+\frac {2m-2}{2m-3}\cdot \color{blue}{(2m-3)}\right)\\\\
&=\frac 1{2m-1}\cdot \color{blue}{(2m-1)}\\\\
&=\color{red}{\large 1}\\\\\end{align}$$

In alternative notation, starting from the innermost parenthesis,
$$\scriptsize\begin{align}
a_1\;\;&=1+\frac 21&&=3\\
a_2\;\;&=1+\frac 43a_1&&=5\\
a_3\;\;&=1+\frac 65 a_2&&=7\\
&\qquad \vdots &\qquad \\
a_n\;\;&=1+\frac {2n}{2n-1}a_{n-1}&&=2n+1\\
&\qquad \vdots &\qquad \\
a_{m-1}&=1+\frac {2m-2}{2m-3}\cdot {(2m-3)}&&=2m-1\\
S\;\;\;&=\frac 1{2m-1}a_{m-1}&&=1
\end{align}$$
