I am trying to show that: \begin{equation} \det\left[ \binom{2n}{n+i-j} \right]_{i,j=0}^{n-1}=\prod_{i=0}^{n-1} \frac{\binom{2n+i}{n}}{\binom{n+i}{n}} \end{equation}
I have tried playing with the algebra for some time. For example, if we fix $i$ and consider a particular row vector, we have: \begin{equation} \left[ \begin{array}{c} \binom{2n}{n+i} & \binom{2n}{n+i-1} & \binom{2n}{n+i-2} & \dots & \binom{2n}{i+1} \end{array} \right] \end{equation} Which equals \begin{equation} \left[ \begin{array}{c} \frac{(2n)!}{(n+i)!(n-i)!} & \frac{(2n)!}{(n+i-1)!(n-i+1)!} & \frac{(2n)!}{(n+i-2)!(n-i+2)!} & \dots & \frac{(2n)!}{(i+1)!(2n-i-1)!} \end{array} \right] \end{equation} It seems that our goal should be to factor out $\binom{2n+i}{n} / \binom{n+i}{n}$ and leave a matrix whose determinant evaluates to 1. Clearly: \begin{equation} \binom{2n+i}{n} / \binom{n+i}{n} = \frac{(2n+i)!}{n!(n+i)!} \cdot \frac{n!(i!)}{(n+i)!} = \frac{(2n+i)!}{(n+i)!} \cdot \frac{i!}{(n+i)!} \end{equation} However, I am unsure of how to proceed.
For those interested, the determinant given enumerates plane partitions contained within an $n\times n \times n$ cube, or equivalently, rhombic tilings of a regular hexagon with side length $n$.