how is a factorial fraction equal to the product notation

Question

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 $2k-3$.

Answer 1

Your product is $1*3*5*7*9...*(2n-3)$.
Multiply and divide by $2*4*6*8... *(2n-4)$.
So, the numerator is $(2n-3)!$.
The denominator uses only the even numbers.
Divide each of them by $2$.
You then see appearing $1*2*3...*(n-2)$ which is $(n-2)!$.
This division by $2$ has been done $(n-2)$ times.
Then ...
Is that clear enough ?