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:

> 
> Does anyone know a *geometric* proof of this result?
> 
> 
> 

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: 
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?
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 $\prod\_{i<j} (j-i)=0!1!\dots (n-1)!$.