I was reading the wikipedia page on the Toom-Cook algorithm, and I found it strange that there are several interpolation matrices given for various $k$ without any mention of how to create interpolation matrices for larger $k$. So, going by the examples provided on the page, I derived a simple algorithm to create similar matrices. While evaluating the determinants of the matrices I generated, (as these matrices must be nonsingular,) I noticed that, for a given $n$, the determinant of the corresponding $n\times n$ matrix is equal to the $(n-2)$th superfactorial! So, my question is this: given the following matrix $M$: $$ M = \begin{pmatrix} 1 & 0 & \cdots & 0 & 0 \\ a_{1,0} & a_{1,1} & \cdots & a_{1,n-2} & a_{1,n-1} \\ \vdots & \vdots & \ddots & \vdots & \vdots \\ a_{n-2,0} & a_{n-2,1} & \cdots & a_{n-2,n-2} & a_{n-2,n-1} \\ 0 & 0 & \cdots & 0 & 1 \end{pmatrix}, $$ where $a_{i,j} = ((-1)^{i+1} \lceil \frac{i}{2}\rceil)^j$, does $|M| = \prod\limits_{k=1}^{n-2}k!$ hold for all $n\in\mathbb{Z}$ and $n>2$?
An example of such a matrix can be seen below:
$$ \begin{pmatrix} 1 & 0 & 0 & 0 & 0 \\ 1 & 1 & 1 & 1 & 1 \\ 1 & -1 & 1 & -1 & 1 \\ 1 & 2 & 4 & 8 & 16 \\ 0 & 0 & 0 & 0 & 1 \end{pmatrix} $$
As the are only one nonzero element in both the first and the last line our determinant is equal determinat of matrix: $$ M' = \begin{pmatrix} a_{1,1} & \cdots & a_{1,n-2} \\ \vdots & \ddots & \vdots \\ a_{n-2,1} & \cdots & a_{n-2,n-2} \\ \end{pmatrix}. $$
For every $i$ element $a_{i,j}$ if the $j$-st degree of the first element of $i$-st line, so $M'$ is Vandermonde matrix multiplied by first column's elements, so $$|M'|=\prod_{i=1}^{n-2} |a_{i,1}| \prod_{1\leq i<j\leq n-2} |a_{i,1}-a_{j,1}|.$$
For $n>3$ we have a recurrence relation $$|M'(n)|=|M'(n-1)||a_{n-2,1}|\prod_{i=1}^{n-3} |a_{i,1}-a_{n-2,1}|=$$ $$=|M'(n-1)|*\lceil \frac{n-2}{2}\rceil*(n-2)*1*(n-3)*2*\dots*\lceil \frac{n}{2}\rceil*\lceil \frac{n-4}{2}\rceil=$$ $$=|M'(n-1)|*(n-2)!,$$ that's just what we need.