Prove: $$ \begin{array}{|cccccccccc|} 1 & 0 & 0 & \cdots & 0 & 1 & 0 & 0 & \cdots & 0 \\ x & x & x & \cdots & x & y & y & y & \cdots & y \\ x^{2} & 2 x^{2} & 2^{2} x^{2} & \cdots & 2^{m-1} x^{2} & y^{2} & 2 y^{2} & 2^{2} y^{2} & \cdots & 2^{m-1} y^{2} \\ x^{3} & 3 x^{3} & 3^{2} x^{3} & \cdots & 3^{m-1} x^{3} & y^{3} & 3 y^{3} & 3^{2} y^{3} & \cdots & 3^{m-1} y^{3} \\ \vdots & \vdots & \vdots & \ddots & \vdots & \vdots & \vdots & \vdots & \ddots & \vdots \\ x^{n} & n x^{n} & n^{2} x^{n} & \cdots & n^{m-1} x^{n} & y^{n} & n y^{n} & n^{2} y^{n} & \cdots & n^{m-1} y^{n} \end{array}=(x-y)^{m^{2}}(x y)^{\frac{m^{2}-m}{2}}\left(\prod_{i=0}^{m-1} i !\right)^{2} $$ where $n=2m-1$.
A friend of mine gave me this problem. He claimed that he solved this by a very complicated method(which is too long to type here, and to be frank, I didn't get it at all). The following part is my progress.
First, we want to extract $y$ so that we can let $z=\frac{x}{y}$, and then the determinant should be a polynomial $f(z)$ of variable $z$. Thus we can use calculus to simplify it. This trick usually works nicely on two-varible homogeneous determinant, but it doesn't kill this problem. Since we know the original problem is equivalent to $$ \begin{array}{|cccccccccc|} 1 & 0 & 0 & \cdots & 0 & 1 & 0 & 0 & \cdots & 0 \\ x & x & x & \cdots & x & 1 & 1 & 1 & \cdots & 1 \\ x^{2} & 2 x^{2} & 2^{2} x^{2} & \cdots & 2^{m-1} x^{2} & 1 & 2 & 2^{2} & \cdots & 2^{m-1} \\ x^{3} & 3 x^{3} & 3^{2} x^{3} & \cdots & 3^{m-1} x^{3} & 1 & 3 & 3^{2} & \cdots & 3^{m-1} \\ \vdots & \vdots & \vdots & \ddots & \vdots & \vdots & \vdots & \vdots & \ddots & \vdots \\ x^{n} & n x^{n} & n^{2} x^{n} & \cdots & n^{m-1} x^{n} & 1 & n & n^{2} & \cdots & n^{m-1} \end{array}=(x-1)^{m^{2}}x^{\frac{m^{2}-m}{2}}\left(\prod_{i=0}^{m-1} i !\right)^{2} $$ Note $f(x)=LHS$, calculate the derivative of $f(x)$, we have $$ x\frac{\,\mathrm{d}}{\,\mathrm{d}x}f(x)=\begin{array}{|cccccccccc|} 1 & 0 & \cdots & 0 & 0 & 1 & 0 & 0 & \cdots & 0 \\ x & x & \cdots & x & x & 1 & 1 & 1 & \cdots & 1 \\ x^{2} & 2 x^{2} & \cdots & 2^{m-2} x^{2} & 2^{m} x^{2} & 1 & 2 & 2^{2} & \cdots & 2^{m-1} \\ x^{3} & 3 x^{3} & \cdots & 3^{m-2} x^{3} & 3^{m} x^{3} & 1 & 3 & 3^{2} & \cdots & 3^{m-1} \\ \vdots & \vdots & \vdots & \ddots & \vdots & \vdots & \vdots & \vdots & \vdots & \ddots \\ x^{n} & n x^{n} & \cdots & n^{m-2} x^{n} & n^{m} x^{n} & 1 & n & n^{2} & \cdots & n^{m-1} \end{array} $$ which doesn't help at all.
I hope you can share some thoughts of this problem.