# Find representation of $\prod_{1 \leq i < j \leq n} (x_i - x_j)$ as a sum of monomials

Of $$\prod_{1 \leq i < j \leq n} (x_i - x_j)$$

Taking into account the matrix $$\begin{pmatrix}1 & x_1 & x_1^2 & \cdots & x_1^n \\ 1& x_2 & x_2^2 & \cdots & x_2^n\\ \vdots & \vdots & \vdots & \ddots & \vdots\\ 1 & x_n & x_n^2 & \cdots & x_n^n\end{pmatrix}$$

I have found that, for $n=3$ the multiplication gives $x_1 x_2 (x_1 - x_2) + x_2x_3(x_2-x_3)+x_1x_3(x_1-x_3)$ which uses all the combinations for $n=3$. Also, I've found that the order of the result will be $n-1$ and that I will get $n!$ monomials. I don't know how to find the answer though.

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Call the Vandermonde determinant:

$$V_n(x_1,x_2,\ldots,x_n)= \left\lvert \begin{matrix} 1 & x_1 & x_1^2 & \cdots & x_1^{n-1} \\ 1& x_2 & x_2^2 & \cdots & x_2^{n-1}\\ \vdots & \vdots & \vdots & \ddots & \vdots\\ 1 & x_n & x_n^2 & \cdots & x_n^{n-1}\end{matrix} \right\rvert$$

Then if we express this via the cofactor expansion of the bottom row it is a polynomial in $x_n$ with $V_{n-1}(x_1,x_2,\ldots,x_{n-1})$ being the coefficient of $x_n^{n-1}$ .

We can also see that replacing $x_n$ with any of $x_1,x_2,\ldots,x_{n-1}$ produces a determinant with two equal rows so they must be the $n$ roots of our polynomial.

Hence we can write the polynomial in terms of its roots

$$V_n(x_1,x_2,\ldots,x_n)=V_{n-1}(x_1,x_2,\ldots,x_{n-1})(x_n-x_1)(x_n-x_2)\ldots(x_n-x_{n-1})$$

then solving the reccurrence by iterating $V_{n-1}(x_1,x_2,\ldots,x_{n-1})$ gives the desired product in the question.

This determinant comes in handy for a proof of the hook length formula for Standard Young's Tableaux.