# Calculating determinant of a Vandermonde type matrix of order n

Det$\begin{bmatrix} 1 & 2 & 3 &\ldots &n\\ 1& 2^3& 3^3& \ldots & n^3\\ 1 &2^5& 3^5& \ldots & n^5\\ \vdots & \vdots& & \vdots \\ 1&2^{2n-1}& 3^{2n-1}& \ldots &n^{2n-1} \end{bmatrix}$

If the powers were consecutively increasing down the rows than we could use the Vandermonde’s identity. However I am quite uncertain about how to tackle this. I am convinced that elementary row operations will be to no avail. Any hints?

If you multiply the second column by $2$, the third by $3$ and so on, you get $$\det \begin{bmatrix} 1 & 2 & 3 &\ldots &n\\ 1& 2^3& 3^3& \ldots & n^3\\ 1 &2^5& 3^5& \ldots & n^5\\ \vdots & \vdots& & \vdots \\ 1&2^{2n-1}& 3^{2n-1}& \ldots &n^{2n-1} \end{bmatrix} = \frac{1}{n!}\det \begin{bmatrix} 1 & 2^2 & 3^2 &\ldots &n^2\\ 1& 2^4& 3^4& \ldots & n^4\\ 1 &2^6& 3^6& \ldots & n^6\\ \vdots & \vdots& & \vdots \\ 1&2^{2n}& 3^{2n}& \ldots &n^{2n} \end{bmatrix}$$ and the matrix is now Vandermonde.
Call $\Delta$ your determinant. It seems to me that $$\Delta = n! \det \left( \begin{array}{ccccc} 1 & 1 & 1 & \dots & 1 \\ 1 & 4 & 9 & \cdots & n^2 \\ 1 & 4^2 & 9^2 & \cdots & (n^2)^2 \\ \vdots & \vdots & \vdots & \cdots & \vdots \\ 1 & 4^{n-1} & 9^{n-1} & \cdots & {(n^2)}^{(n-1)} \end{array} \right) = n! \left( \begin{array}{ccccc} 1 & 1 & 1 & \dots & 1 \\ x_1 & x_2 & x_3 & \cdots & x_n \\ x_1^2 & x_2^2 & x_3^2 & \cdots & x_n^2 \\ \vdots & \vdots & \vdots & \cdots & \vdots \\ x_1^{n-1} & x_2^{n-1} & x_3^{n-1} & \cdots & x_n^{n-1} \end{array} \right)$$ were $x_1 = 1, x_2=4, \cdots, x_i = i^2 , \cdots, x_n = n^2$. From which you can use the usual Vandermonde.