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Compute the determinant of the matrix follows, $$\begin{pmatrix}1&2&\cdots&n\\1^2&2^2&\cdots&n^2\\\vdots&\vdots&\ddots&\vdots\\1^n&2^n&\cdots&n^n\end{pmatrix}$$

Thanks in advance.

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    $\begingroup$ Scale the columns by $\frac11, \frac12, \ldots\frac1n$, and what you have left is exactly a Vandermonde determinant times $n!$. $\endgroup$ – Henning Makholm Sep 17 '15 at 12:44
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    $\begingroup$ Is $a_{2n}=2^n$ a typo? $\endgroup$ – Zoran Loncarevic Sep 17 '15 at 12:48
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It is Vandermonde for $x_i=i$ can be also written as $$\prod_{k=1}^{n+2} (k-1)! $$ there is a special function for that called Barnes G-function. In terms of it $$ G(n+2) $$

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