A determinant of perfect square polynomials Usually, I like working with determinants related to the Vandermonde matrix, i.e.
$$\det(x_j^{i-1})=\prod_{i<j}(x_j-x_i).$$
However, I run into some unusual matrix and its determinant. Define the $(2n)\times (2n)$ matrix $\mathbb{M}_{2n}$ with entries
$$\mathbb{M}_{2n}(i,j)=
\begin{cases}
(x_j-x_i)(x_i-x_{2n}) \qquad \text{if $i<j$}, \\
(x_j-x_i)(x_j-x_{2n}) \qquad \text{if $i\geq j$}.
\end{cases}$$

QUESTION. Is this true?
$$\det\mathbb{M}_{2n}=(x_1-x_2)^2(x_2-x_3)^2\cdots(x_{2n-1}-x_{2n})^2(x_{2n}-x_1)^2.$$

 A: The answer is yes for any $n \geq 2$. (The $n=1$ case should be treated separately.)
The determinant, as a polynomial, contains $(x_m−x_{m+1})$ $(m=1,2,...,2n-1)$ and $(x_{2n}-x_1)$ as factors, because setting $x_m=x_{m+1}$ will make the $m$th and $(m+1)$th column of $\mathbb{M}_{2n}$ equal, and setting $x_{2n}=x_1$ will make the $(2n)$th and first column of $\mathbb{M}_{2n}$ equal, making the determinant zero.
Since the determinant of a skew-symmetric matrix can always be written as the square of a polynomial in the matrix entries that only depend on the size of the matrix, the determinant contains the factor $(x_1−x_2)^2(x_2−x_3)^2⋯(x_{2n−1}−x_{2n})^2(x_{2n}−x_1)^2$. As the determinant cannot have degree higher than that of this factor, the only thing left is to plug in values of $x_k$ such that LHS=RHS≠0 in the last equation stated in the question.
If we let $x_k=1$ if $k$ is odd and $0$ otherwise, then
$\mathbb{M}_{2n}(i,j)=
\begin{cases}
\phantom{-}1 \qquad  \text{if $i<j$},i \text{ odd}, j \text{ even} \\
-1 \qquad \text{if $i > j$}, i \text{ even}, j \text{ odd}\\
\phantom{-}0 \qquad  \text{otherwise}
\end{cases}$.
The $(2n-1)$th row has only one nonzero element, namely $\mathbb M_{2n-1,2n}=1$, and the $(2n-1)$th column has only one nonzero element, namely $\mathbb M_{2n,2n-1}=-1$. By applying the Leibniz formula for determinants, we see that $\det \mathbb M_{2n}=\det \mathbb M_{2n-2}$. By mathematical induction, we have $\det \mathbb{M}_{2n}=1$ for any $n$. As the RHS also evaluates to $1$, we claim that LHS=RHS for any $x_k$s.
