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    I need to show the determinant of the Vandermonde matrix is \displaystyle \prod_{1\leq i < j \leq n} (a_j - a_i).

    Now I have looked at the wiki page on this and what I don't get is how

    \displaystyle \prod_{1\leq i < j \leq n} (a_j - a_i) = \sum_{\rho \in Sym(n)} sign(\rho) \prod^n_{i=1} a_{i \ \rho i}

    I think this comes from my not knowing what it means to have the product of the two different dummy variables, i and j.
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    It means you take every possible pair of (i, j) with 1<=i < j <=n and multiply the (a_j - a_i) terms you get as a result.

    e.g. if n=3, the possible terms are (a_2-a_1), (a_3 - a_1) and (a_3 - a_2) and so the product is
    (a_2-a_1)(a_3-a_1)(a_3-a_2).
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    Ok, it turns out that wasn't the problem. How would you go about changing that to a sum of products.

    I think I need to show:

    \displaystyle \prod_{1\leq i &lt; j \leq n} (a_j - a_i) = \sum_{\rho \in Sym(n)} sign(\rho) \prod^n_{i=1} a_i^{\rho i - 1}

    But I am having trouble on getting from one to the other.

    I think I should have j=pi but im not sure how this helps me
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    You're looking at this the wrong way. The sum you have comes directly from the definition of the determinant of a matrix. It's not going to help you at all here.

    You need to work directly from what you know about determinants:

    You know that if two columns of the matrix are the same, the determinant is zero. But if 1 <= i < j <=n and a_i = a_j, then two columns are the same. It follows that (a_j-a_i) is a factor of the determinant for each such pair. By comparing the degree of the product with the degree of the determinant, you can see there are no other factors. At this point you know that the determinant is some constant multiple of the product. By comparing coefficients, you can show the constant is one, and then you're done.
 
 
 
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