I have a textbook (Asymptopia by Joel Spencer, p.66) that states that
$$\frac{(n/2)!(n/2)!}{\left(\frac{n+i}2\right)!\left(\frac{n-i}2\right)!} =\prod_{j=1}^{i/2}\frac{\frac n2+1-j}{\frac n2+j}.$$
The equivalence between the LHS and $\left(\prod_{j=1}^{n/2}j^2\right)\left(\prod_{j=1}^{(n+i)/2}j^{-1}\right)\left(\prod_{j=1}^{(n-i)/2}j^{-1}\right)$ I understand, it results from the product notation form of the factorial and some simple algebraic manipulation. However, for the life of me, I don't see how you can then manipulate it to get the form of the RHS. I've numerically tested it out with different values, so I believe that it is true. For example, if $n=20$ and $i=2$, all terms equal $10/11$. Could someone help explain how one arrives at the RHS expression?
Edit: I believe that the book implies that, due to the use of $n$ even, or the floor of $n/2$, $i$ should be even.
Assume that $n$ and $i$ are even. Then,
$$\frac{(n/2)!(n/2)!}{\left(\frac{n+i}2\right)!\left(\frac{n-i}2\right)!}=\frac AB$$ where $$A=\frac{(n/2)!}{\left(\frac{n-i}2\right)!}=\prod_{k=\frac{n-i}2+1}^{n/2}k=\prod_{j=1}^{i/2}\left(\frac n2+1-j\right)$$ and $$B=\frac{\left(\frac{n+i}2\right)!}{(n/2)!}=\prod_{k=\frac n2+1}^{(n+i)/2}k=\prod_{j=1}^{i/2}\left(\frac n2+j\right).$$