This question came up in a recent problem. It basically states: In the limit of $N\gt\gt m \gt\gt 1$ show that $$\frac{\frac{N}{2}!\frac{N}{2}!}{(\frac{N}{2}-m)!(\frac{N}{2}+m)!} \approx \exp\left({-\frac{2m^2}{N}}\right)$$
I'm able to show this approximation by taking the logarithm of the left hand side, using stirling's approximation on the factorials, keeping the lowest order terms and taking the exponential again.
However I would like to show this approximation holds without taking the logarithm and using stirlings approximation.
Simple canceling of common factors gives the expression: $$\frac{\left(\frac{N}{2}\right){}\left(\frac{N}{2}-1\right)...\left(\frac{N}{2}-(m-1)\right)}{\left(\frac{N}{2}+m\right)\left(\frac{N}{2}+(m-1)\right)...\left(\frac{N}{2}+1\right)}$$
Grouping similar terms gave me the following expression: $$\frac{1}{1+\frac{2m}{N}}\prod_{i=1}^{m-1}{\frac{1-\frac{2(m-i)}{N}}{1+\frac{2(m-i)}{N}}}$$
However this is the point where I'm stuck. I tried using binomial approximations on the leading term and the terms in the product and ignoring higher order terms, but I didn't get anything resembling a product/sum approximating an exponential.
Any help would be greatly appreciated.