Evaluate the determinant $\det\left[ \binom{2n}{n+i-j} \right]_{i,j=0}^{n-1}$ 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 $n$.
 A: Krattenthaler, in this article, proves a more general formula, of which the OP's determinant is a special case. Given the $n\times n$ matrix $\mathbf A$ with elements
$$\mathbf a_{j,k}=\binom{p+q}{p+j-k}, \quad 1\leq j,k\leq n$$
then
$$\begin{align*}
\det\mathbf A&=\prod_{j=1}^n \prod_{k=1}^p \prod_{\ell=1}^q \frac{j+k+\ell-1}{j+k+\ell-2}\\
&=\prod_{j=1}^n \frac{(p+q+j-1)!(j-1)!}{(p+j-1)!(q+j-1)!}
\end{align*}$$
where the triple product formula is attributed to MacMahon. A number of proofs for this determinantal identity are given in the linked article. For your particular special case,
$$\prod_{j=0}^{n-1} \frac{(2n+j)!j!}{((n+j)!)^2}=\prod_{j=0}^{n-1} \frac{\frac{(2n+j)!}{(n+j)!n!}}{\frac{(n+j)!}{n!j!}}=\prod_{j=0}^{n-1} \frac{\binom{2n+j}{n}}{\binom{n+j}{n}}$$
See this article as well.
A: 
It seems that our goal should be to factor out ... and
  leave a matrix whose determinant evaluates to 1.

Actually, one can factor out something to leave a (generalized) Vandermonde matrix.
Using column operations one can see that
$$
\det\binom{2n}{n+i-j}=\det\binom{2n+j}{n+i}.
$$
Now $\binom{n}{k}=\frac{n^{\downarrow k}}{k!}$, where $n^{\downarrow k}:=n(n-1)\ldots(n-k+1)$, so the determinant we want to compute is just
$$
\det\frac{(2n+j)^{\downarrow n+i}}{(n+i)!}=
\prod_i\frac1{(n+i)!}\cdot\prod_j(2n+j)^{\downarrow n}\cdot \det(n+j)^{\downarrow i}
$$
(here we used that $a^{\downarrow k+l}=a^{\downarrow k}\cdot(a-k)^{\downarrow l}$).
Observe that
$$
\det(n+j)^{\downarrow i}=\det(n+j)^i=\prod j!,
$$
so the answer is
$$
\prod_j\frac{(2n+j)!j!}{((n+j)!)^2}=
\prod_j\frac{\binom{2n+j}n}{\binom{n+j}n}.
$$
(Perhaps, all this is also contained in Krattenthaler's paper quoted above. But anyway.)
