<s>[INST] 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?
 [/INST] 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.
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