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?
 A: 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.
A: 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.
