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?
 A: HINT:
Note that we can write
$$1\cdot3\cdot 5\cdots (2k-1)=\frac{1\cdot 2\cdot 3\cdot 4\cdots (2k-2)(2k-1)(2k)}{2(1)\cdot 2(2)\cdots 2(k-1)2(k)}=\frac{(2k)!}{2^k\,k!}$$
A: $$\prod_{r=1}^k(2r-1)=\dfrac{\prod_{r=1}^{2k}r}{2^k\cdot k!}=?$$
A: \begin{align}
\frac{1\cdot3\cdot5\cdot7\cdot11\cdot13}{2^7\cdot2!} & = \frac{1\cdot2\cdot3\cdot4\cdot5\cdot6\cdot7\cdot8\cdot9\cdot10\cdot11\cdot12\cdot13\cdot14}{2^7 \cdot2\cdot4\cdot6\cdot8\cdot10\cdot12\cdot14} \\[10pt]
& = \frac{14!}{2^7\cdot(2\cdot2\cdot2\cdot2\cdot2\cdot2\cdot2)\cdot(1\cdot2\cdot3\cdot4\cdot5\cdot6\cdot7)} = \frac{14!}{2^7\cdot2^7\cdot7!}
\end{align}
A: 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.
