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.
This isn't the answer you're looking for, but I discovered it while attempting to find a geometric approach. 

Consider the polynomials $$p\_j(x) = (x-x\_1)\cdots (x-x\_j) = \sum\_{i=0}^{n-1} a\_{i,j} x^i,$$
where $p\_0(x)=1$ by convention, $0\leq j\leq n-1$. 
Of course, $a\_{j,j}=1$ and $a\_{i,j}=0$ if $i>j$, and $a\_{i,j}$ is the $i$th symmetric polynomial in $x\_1,\ldots, x\_j$ up to sign. 
If we multiply the Vandermonde matrix $[x\_i^{j-1}]$ by the upper unipotent matrix $[a\_{i-1,j-1}]$, 
we get the matrix $[p\_{j-1}(x\_i)]$. This is a lower triangular matrix, since $p\_{j-1}(x\_i) =0$ if $i \leq j-1$, and the diagonal entries are $p\_{i-1}(x\_i)$. 
Clearly, 
$$\prod\_{i=1}^n p\_{i-1}(x\_i) = \prod\_{i=1}^n (x\_i-x\_1) \cdots (x\_i-x\_{i-1}) = \prod\_{1\leq i < j\leq n} (x\_j-x\_i).$$

Maybe there's a way of seeing this product geometrically as a natural affine transformation?