Evaluate the determinant $\det\left[ \binom{2n}{n+i-j} \right]_{i,j=0}^{n-1}$

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.

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.)