<s>[INST] Why is $1 \times 3 \times 5 \times \cdots \times (2k-3) = \frac{(2k-2)!}{2^{(k-1)}(k-1)!}$ In order to find out the Catalan numbers from their generating function you have to evaluate the product above.
Here is what I thought:
\begin{align*}
1 \times 3 \times 5 \times...\times  (2k-3) &= \frac{1 \times 2 \times 3 \times 4 \times \cdots \times (2k - 3)}{2 \times 4 \times \cdots \times (2k - 2)} \\
&= \frac{(2k - 3)!}{2^{k - 1}(k - 1)!} \\
\end{align*}
But a Mathematica session quickly proved me wrong, instead the result in the title is true. Why is that true, what mistake did I make?
 [/INST] For example, for the denominator $$2\times 4\times 6\times 8=2\cdot 1\times 2\cdot 2\times 2\cdot 3\times 2\cdot 4=2\times 2\times 2\times 2\cdot 1\times 2\times 3\times 4=2^4\cdot 4!$$ 
I let you adapt this hint to your exercise.
</s>