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Denis Serre
Denis Serre
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Surely, there is nothing special in the all-ones vector: the claim holds for any integer-valued $e$.

Notice that $$ \det W^TW =\det\bigl[e^T (-1)^iS^{I+j}e\bigr], $$$$ \det W^TW =\det\bigl[e^T (-1)^iS^{i+j}e\bigr], $$ Since $S$ is skew-symmetric, we have $e^TS^ie=0$ for all odd $i$. Permuting now the rows and columns of $W^TW$ we get $$ \det W^TW=\det\begin{bmatrix} A_0&0\\ 0& -A_1\end{bmatrix}, $$ where $$ A_0= \begin{bmatrix} e^Te& e^TS^2e& e^TS^4e& \cdots& e^TS^{n-2}e\\ e^TS^2e& e^TS^4e& e^TS^6e& \cdots& e^TS^ne\\ \vdots& \vdots& \vdots& \ddots& \vdots\\ e^TS^{n-2}e& e^TS^ne& e^TS^{n+2}e& \cdots& e^TS^{2n-4}e \end{bmatrix}, \quad A_1= \begin{bmatrix} e^TS^2e& e^TS^4e& e^TS^6e& \cdots& e^TS^ne\\ e^TS^4e& e^TS^6e& e^TS^8e& \cdots& e^TS^{n+2}e\\ \vdots& \vdots& \vdots& \ddots& \vdots\\ e^TS^{n}e& e^TS^{n+2}e& e^TS^{n+4}e& \cdots& e^TS^{2n-2}e \end{bmatrix}. $$ We show that $$ \det A_1=\pm\det S\det A_0, \qquad(*) $$ so that $\det W=\pm\det A_0\sqrt{\det S}$, which yields what we want.

Denote $$ u_i= \begin{bmatrix} e^TS^ie& e^TS^{i+2}e& \cdots& e^TS^{i+n-2}e \end{bmatrix}. $$ The eigenvalues of $S$ are purely imaginary, so by Cayley—Hamilton we get $$ S^{n}=\sum_{j=0}^{n/2-1} \alpha_jS^{2j}, $$ where $\alpha_0=-\det S$. Therefore, $$ u_n= -\det S\, u_0+\sum_{j=1}^{n/2-1} \alpha_ju_{2j}. $$ Plug this into $\det A_1$, expand by linearity, and erase vanishing summands to get $(*)$.

Surely, there is nothing special in the all-ones vector: the claim holds for any integer-valued $e$.

Notice that $$ \det W^TW =\det\bigl[e^T (-1)^iS^{I+j}e\bigr], $$ Since $S$ is skew-symmetric, we have $e^TS^ie=0$ for all odd $i$. Permuting now the rows and columns of $W^TW$ we get $$ \det W^TW=\det\begin{bmatrix} A_0&0\\ 0& -A_1\end{bmatrix}, $$ where $$ A_0= \begin{bmatrix} e^Te& e^TS^2e& e^TS^4e& \cdots& e^TS^{n-2}e\\ e^TS^2e& e^TS^4e& e^TS^6e& \cdots& e^TS^ne\\ \vdots& \vdots& \vdots& \ddots& \vdots\\ e^TS^{n-2}e& e^TS^ne& e^TS^{n+2}e& \cdots& e^TS^{2n-4}e \end{bmatrix}, \quad A_1= \begin{bmatrix} e^TS^2e& e^TS^4e& e^TS^6e& \cdots& e^TS^ne\\ e^TS^4e& e^TS^6e& e^TS^8e& \cdots& e^TS^{n+2}e\\ \vdots& \vdots& \vdots& \ddots& \vdots\\ e^TS^{n}e& e^TS^{n+2}e& e^TS^{n+4}e& \cdots& e^TS^{2n-2}e \end{bmatrix}. $$ We show that $$ \det A_1=\pm\det S\det A_0, \qquad(*) $$ so that $\det W=\pm\det A_0\sqrt{\det S}$, which yields what we want.

Denote $$ u_i= \begin{bmatrix} e^TS^ie& e^TS^{i+2}e& \cdots& e^TS^{i+n-2}e \end{bmatrix}. $$ The eigenvalues of $S$ are purely imaginary, so by Cayley—Hamilton we get $$ S^{n}=\sum_{j=0}^{n/2-1} \alpha_jS^{2j}, $$ where $\alpha_0=-\det S$. Therefore, $$ u_n= -\det S\, u_0+\sum_{j=1}^{n/2-1} \alpha_ju_{2j}. $$ Plug this into $\det A_1$, expand by linearity, and erase vanishing summands to get $(*)$.

Surely, there is nothing special in the all-ones vector: the claim holds for any integer-valued $e$.

Notice that $$ \det W^TW =\det\bigl[e^T (-1)^iS^{i+j}e\bigr], $$ Since $S$ is skew-symmetric, we have $e^TS^ie=0$ for all odd $i$. Permuting now the rows and columns of $W^TW$ we get $$ \det W^TW=\det\begin{bmatrix} A_0&0\\ 0& -A_1\end{bmatrix}, $$ where $$ A_0= \begin{bmatrix} e^Te& e^TS^2e& e^TS^4e& \cdots& e^TS^{n-2}e\\ e^TS^2e& e^TS^4e& e^TS^6e& \cdots& e^TS^ne\\ \vdots& \vdots& \vdots& \ddots& \vdots\\ e^TS^{n-2}e& e^TS^ne& e^TS^{n+2}e& \cdots& e^TS^{2n-4}e \end{bmatrix}, \quad A_1= \begin{bmatrix} e^TS^2e& e^TS^4e& e^TS^6e& \cdots& e^TS^ne\\ e^TS^4e& e^TS^6e& e^TS^8e& \cdots& e^TS^{n+2}e\\ \vdots& \vdots& \vdots& \ddots& \vdots\\ e^TS^{n}e& e^TS^{n+2}e& e^TS^{n+4}e& \cdots& e^TS^{2n-2}e \end{bmatrix}. $$ We show that $$ \det A_1=\pm\det S\det A_0, \qquad(*) $$ so that $\det W=\pm\det A_0\sqrt{\det S}$, which yields what we want.

Denote $$ u_i= \begin{bmatrix} e^TS^ie& e^TS^{i+2}e& \cdots& e^TS^{i+n-2}e \end{bmatrix}. $$ The eigenvalues of $S$ are purely imaginary, so by Cayley—Hamilton we get $$ S^{n}=\sum_{j=0}^{n/2-1} \alpha_jS^{2j}, $$ where $\alpha_0=-\det S$. Therefore, $$ u_n= -\det S\, u_0+\sum_{j=1}^{n/2-1} \alpha_ju_{2j}. $$ Plug this into $\det A_1$, expand by linearity, and erase vanishing summands to get $(*)$.

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Ilya Bogdanov
Ilya Bogdanov
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Surely, there is nothing special in the all-ones vector: the claim holds for any integer-valued $e$.

Notice that $$ \det W^TW =\det\bigl[e^T (-1)^iS^{I+j}e\bigr], $$ Since $S$ is skew-symmetric, we have $e^TS^ie=0$ for all odd $i$. Permuting now the rows and columns of $W^TW$ we get $$ \det W^TW=\det\begin{bmatrix} A_0&0\\ 0& -A_1\end{bmatrix}, $$ where $$ A_0= \begin{bmatrix} e^Te& e^TS^2e& e^TS^4e& \cdots& e^TS^{n-2}e\\ e^TS^2e& e^TS^4e& e^TS^6e& \cdots& e^TS^ne\\ \vdots& \vdots& \vdots& \ddots& \vdots\\ e^TS^{n-2}e& e^TS^ne& e^TS^{n+2}e& \cdots& e^TS^{2n-4}e \end{bmatrix}, \quad A_1= \begin{bmatrix} e^TS^2e& e^TS^4e& e^TS^6e& \cdots& e^TS^ne\\ e^TS^4e& e^TS^6e& e^TS^8e& \cdots& e^TS^{n+2}e\\ \vdots& \vdots& \vdots& \ddots& \vdots\\ e^TS^{n}e& e^TS^{n+2}e& e^TS^{n+4}e& \cdots& e^TS^{2n-2}e \end{bmatrix}. $$ We show that $$ \det A_1=\pm\det S\det A_0, \qquad(*) $$ so that $\det W=\pm\det A_0\sqrt{\det S}$, which yields what we want.

Denote $$ u_i= \begin{bmatrix} e^TS^ie& e^TS^{i+2}e& \cdots& e^TS^{i+n-2}e \end{bmatrix}. $$ The eigenvalues of $S$ are purely imaginary, so by Cayley—Hamilton we get $$ S^{n}=\sum_{j=0}^{n/2-1} \alpha_jS^{2j}, $$ where $\alpha_0=-\det S$. Therefore, $$ u_n= -\det S\, u_0+\sum_{j=1}^{n/2-1} \alpha_ju_{2j}. $$ Plug this into $\det A_1$, expand by linearity, and erase vanishing summands to get $(*)$.