Factorial equality $\ \frac{1\cdot3\cdot5\cdots(2k-1)}{2^k2!}$ $\ = \frac{(2k)!}{2^k2^kk!k!}$ In a generating function identity proof in my textbook there is a step that I can't wrap my head around.
$$ \frac{1\cdot3\cdot5\cdots(2k-1)}{2^k2!}$$
$$ = \frac{(2k)!}{2^k 2^k k!k!}$$
How does one get from the left side of the equation to the right side? Is there an intuitive explanation as for why this makes sense?

In the denominator of the lhs, $2!$ should be $k!$
Let $1\cdot 3\cdot 5\cdot \cdot \cdot (2k-1) = (2k-1)!!\,$, the double factorial.
We want to show that
\begin{equation}
(2k-1)!! = \frac{(2k)!}{2^{k}k!}
\end{equation}
We have
\begin{align}
(2k-1)!!2^{k}k! &= [(2k-1)(2k-3)\cdot \cdot \cdot 1]2^{k}k(k-1)(k-2)\cdot \cdot \cdot 1 \\
\tag{1}
&= [(2k-1)(2k-3)\cdot \cdot \cdot 1][2k][2(k-1)][2(k-2)]\cdot \cdot \cdot [2\cdot 1] \\
\tag{2}
&= 2k(2k-1)(2k-2)\cdot \cdot \cdot 1\\
&= (2k)!
\end{align}


*

*Associate each $2$ from $2^{k}$ with each term of $k!$

*Rearrange terms.
