# Factorial equality (1*3*5*(2k-1))/(2^k2!}=((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. 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?

Factorial equality
In a generating function identity proof in my textbook there is a step that I can't wrap my head around.
$\frac{1\cdot 3\cdot 5\cdots \left(2k-1\right)}{{2}^{k}2!}$
$=\frac{\left(2k\right)!}{{2}^{k}{2}^{k}k!k!}$
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?
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Royce Golden
HINT:
Note that we can write
$1\cdot 3\cdot 5\cdots \left(2k-1\right)=\frac{1\cdot 2\cdot 3\cdot 4\cdots \left(2k-2\right)\left(2k-1\right)\left(2k\right)}{2\left(1\right)\cdot 2\left(2\right)\cdots 2\left(k-1\right)2\left(k\right)}=\frac{\left(2k\right)!}{{2}^{k}\phantom{\rule{thinmathspace}{0ex}}k!}$
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Landen Miller
$\begin{array}{rl}\frac{1\cdot 3\cdot 5\cdot 7\cdot 11\cdot 13}{{2}^{7}\cdot 2!}& =\frac{1\cdot 2\cdot 3\cdot 4\cdot 5\cdot 6\cdot 7\cdot 8\cdot 9\cdot 10\cdot 11\cdot 12\cdot 13\cdot 14}{{2}^{7}\cdot 2\cdot 4\cdot 6\cdot 8\cdot 10\cdot 12\cdot 14}\\ & =\frac{14!}{{2}^{7}\cdot \left(2\cdot 2\cdot 2\cdot 2\cdot 2\cdot 2\cdot 2\right)\cdot \left(1\cdot 2\cdot 3\cdot 4\cdot 5\cdot 6\cdot 7\right)}=\frac{14!}{{2}^{7}\cdot {2}^{7}\cdot 7!}\end{array}$
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