<s>[INST] 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?
 [/INST] 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.
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