Question: The Vandermonde matrix is the $n\times n$ matrix whose $(i,j)$-th component is $x\_j^{i-1}$, where the $x\_j$ are indeterminates. It is well known that the determinant of this matrix is $$\prod\_{1\leq i < j \leq n} (x\_j-x\_i).$$

There are many known proofs of this fact, using for example row reduction or the Laplace expansion ([here](http://www.proofwiki.org/wiki/Vandermonde_Determinant)), a combinatorial proof by Art Benjamin and Gregory Dresden ([here](http://home.wlu.edu/~dresdeng/papers/VDM.pdf)), and another (slightly less) combinatorial proof by Jennifer Quinn ([here](http://www.sciencedirect.com/science/article/pii/S0378375810000406), unfortunately not open access). An easy proof follows by noting that the variety of the determinant contains (as a set) the variety of $x\_i-x\_j$ for all $i < j$ and then by computing the degree of the determinant as a polynomial in the $x\_i, x\_j$, though I don't know a reference for this proof.

Given that this result is amenable to such a wide variety of proofs (the above list contains three somewhat different flavors of proof---linear algebraic, combinatorial, and algebra-geometric), I have the following question:

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> Does anyone know a *geometric* proof of this result?
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For example, one might compute the volume of the parallelepiped whose vertices are given by the rows or columns of this matrix in a clever way. Ideally this would not just boil down to row reduction.

Answer: 
First, I'll prove a lemma: on the rational normal curve of degree $n-1$ in $\mathbb{P}^{n-1}$, any collection of $n$ points spans the whole projective space. This is basically a consequence of the notion of degree: assume not. Then all $n$ points are contained in a hyperplane, but the curve is degree $n-1$, which means that no hyperplane can contain MORE than $n-1$ points, contradiction.

Thus, $n$ points on the rational normal curve are linearly independent, and so the determinant will vanish if and only if two or more of the points coincide. Using this, we can see that the determinant is a scalar multiple of the desired polynomial. Then, to see that the scalar is one, we just pick a coefficient on each side and compare.
My favorite proof is the following (I am not sure that it is geometric, but however). At first, replace each $x\_i^k$ to $x\_i(x\_i-h)(x\_i-2h)\dots (x\_i-(k-1)h)$. I claim that determinant still equals $\prod (x\_j-x\_i)$. Then for $h=0$ we get what we need, while the trick is that we prove it for any other $h$ instead. Say, for $h=1$, it suffices by homogenuity. Note that both determinant and product of differences are polynomials of degree (say, at most) $n(n-1)/2$. It suffices to check that their values coincide on the set of points $\{(x\_1,x\_2,\dots,x\_n)=(c\_1,c\_2,\dots,c\_n)\}$, where $c\_i$ are non-negative integers and $\sum c\_i\leq n(n-1)/2$. This is easy and useful lemma. And in those points everything is clear. Both sides do vanish unless $(c\_1,\dots,c\_n)$ is a permutation of $0,\dots,n-1$. In this latter case the determinant contains unique non-zero summand, and it just equals the product of the sign of above permutation and $\prod\_{i<j} (j-i)=0!1!\dots (n-1)!$.