Q:

Simplify the series given by the recurrence relation $na_n=2a_{n-2}$

If you are given a recurrence relation such that:
$$na_n=2a_{n-2}\implies a_n= \begin{cases}
0  & \text{odd} \,n \\
\frac{2}{n}a_{n-2} & \text{even} \,n
\end{cases}$$
My textbook suggests that the series can be simplified by

Putting $n=2m$ (since only even terms appear in this series), we get $$a_{2m}=\frac{2}{2m}a_{2m-2}=\bbox[#AF0]{\frac{1}{m}a_{2m-\color{red}{2}}=\frac{1}{m}\color{red}{\frac{1}{(m-1)}}a_{2m-\color{red}{4}}}=\frac{1}{m!}a_0$$

I understand the final equality as $$\frac{1}{m}\frac{1}{(m-1)}\frac{1}{(m-2)}\frac{1}{(m-3)}\cdots=\frac{1}{m!}$$
But I do not understand the highlighted equality.
Why does the $\color{red}{\cfrac{1}{m-1}}$ appear when $a_{2m-\color{red}{2}}$ is reduced to $a_{2m-\color{red}{4}}$?

A:

From
$$
a_n=\frac{2}{n}\:a_{n-2},\quad n\,\, \text{even}
$$ you may just put $n=2m-2$ to obtain
$$
a_{2m-\color{red}{2}}=\color{blue}{\frac{2}{2m-2}}\:a_{2m-\color{red}{2-2}}
$$ or
$$
a_{2m-\color{red}{2}}=\color{blue}{\frac1{m-1}}\:a_{2m-\color{red}{4}}.
$$

