Fix an integer $m \geq 2$ and consider the following square matrix with entries in $\Bbb{R}[x_0,\dotsc,x_m]$: $$ M_m = \begin{pmatrix} 1 & x_0 & x_0^2 & \cdots & x_0^{m+1} \\ \vdots & & & & \vdots \\ 1 & x_m & x_m^2 & \cdots & x_m^{m+1} \\ 0 & 1 & 2 x_0 & \cdots & (m+1) x_0^m \end{pmatrix}. $$ Factoring the determinant for small values of $m$ via row operations suggests that $$ \left\lvert \det M_m \right\rvert = \left( \prod_{i = 1}^{m} (x_i - x_0)^2 \right) \left( \prod_{0 < i < j \leq m} \left\lvert x_i - x_j \right\rvert \right), $$ however I am having trouble proving that this is the case in general. Is there a way to prove or disprove this claim?
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Schendel's theorem is what you want. See https://math.stackexchange.com/questions/3743131/closed-form-solution-for-the-determinant-of-a-vandermonde-like-matrix/ – kimchi lover Aug 11 '20 at 14:31
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@kimchilover Thanks, that's exactly what I needed. I am happy for this question to be closed as a duplicate. – A.P. Aug 11 '20 at 15:33
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You don't need any big machinery: just write down the usual Vandermonde determinant in variables $x_0,\dots, x_{m+1}$, differentiate wrt $x_{m+1}$ and then substitute $x_{m+1}:=x_{0}$. The answer drops out - and you get the determinant, not its modulus. – ancient mathematician Aug 11 '20 at 15:33