<s>[INST] how is a factorial fraction equal to the product notation How is the $\prod_{k=2}^n(2k-3)={(2n-3)!\over 2^{n-2}(n-2)!}$, where $n \geq 2$
Note:
I know that the $(2n-3)!$ is equal to the product of $2k-3$ from $k=2$ to $n$, but I can't figure out the bottom half of how diving by $2^{n-2}(n-2)!$ equals the product notation of $2k-3$.
 [/INST] Your equation $\prod_{k=2}^n(2k-3)$ can be expanded to $(2n-3)(2n-5)\cdots(3)(1)$. Clearly, we have every other term for (2n-3)! since we only have the odd values.
Hence,$(2n-3)! = (2n-3)(2n-4)\cdots(2)(1) = (2n-4)(2n-6)\cdots(4)(2) \prod_{k=2}^n(2k-3)$.
Taking the addition terms on the RHS, $(2n-4)(2n-6)\cdots(4)(2) = 2^{n-2}[(n-2)(n-3)\cdots(2)(1)] = 2^{n-2}(n-2)!$
Thus, $(2n-3)! = \prod_{k=2}^n(2k-3) 2^{n-2}(n-2)!$, which you simplify to get the desired result.
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