I'm probably having a brain-fart but I can't figure out why this identity holds:
$$ \prod_{k=0}^{n-1} (n^2-k^2) = \frac{(2n)!}{2} $$
I tried using various formulae involving $\binom{2n}{n}$, without success. Any ideas?
Bonus if it helps visualise why the factors on the RHS, involving squares, would multiply to get the RHS, which does not contain any square terms.
A variation: We obtain \begin{align*} \color{blue}{\prod_{k=0}^{n-1}\left(n^2-k^2\right)} &=\prod_{k=0}^{n-1}(n-k)\prod_{k=0}^{n-1}(n+k)\\ &=\prod_{k=0}^{n-1}(k+1)\cdot n\cdot\prod_{k=1}^{n-1}(n+k)\tag{1}\\ &=n!\cdot n\cdot \frac{(2n-1)!}{n!}\\ &\,\,\color{blue}{=\frac{1}{2}(2n)!} \end{align*} and the claim follows.
In (1) we change the order of multiplication in the left-hand product $k\to n-1-k$ and split the factor $n$ from the right-hand product