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: 
A proof that may be called "geometric" with some good will from your side is as follows. Denote
$V(a\_1,a\_2,\dots,a\_n)$ the Vandermonde matrix with entries $a\_j^{\ i-1}$, and consider the function 
$$f(t):=\det V(a\_1+t, a\_2+t,\dots, a\_n+t \,).$$ One easily computes the derivative of the matrix and notices that $$\partial\_t V:= NV,$$ for a certain matrix $N$ with null trace (actually, a nilpotent matrix with at most $n-1$ non-zero entries). Therefore by the Liouville formula, $$f^{\\ '}(t)=\mathrm{tr}(N)
\ f(t)=0,$$ 
that proves that the Vandermonde determinant is translation-invariant. In particular for $t:=-a\_1$
one has a reduction step that leads plainly to the product formula. (Alternatively, one can solve the above linear ODE getting $V=\exp(tN)V(0)$ and conclude as above, for $\exp(tN)$ is a triangular matrix with unit diagonal entries, hence with determinant equal to 1).
Taking out constants, the Vandermonde determinant is the jacobian of the map $$x\mapsto(x\_1+\dots+x\_n,\dots,x\_1^n+\dots+x\_n^n)$$ (not zero since the first $n$ power sums are algebraically independent). By a classical result on Finite Reflection Groups (R. Steinberg, Invariants of finite reflection groups, Canad. J. Math. 12 (1960), 616–618.; for a nice geometric proof see Humphreys' book: Reflection Groups and Coxeter Groups, section 3.13), this determinant is, up to multiplication by a nonzero constant, the product of linear forms corresponding to the hyperplanes of reflection associated to the reflection group with basic invariants: $x\_1+\dots+x\_n,\dots,x\_1^n+\dots+x\_n^n$ (in this case it is the symmetric group $S\_n$ acting on $\mathbb{R}^n$ with the reflections that send $e\_i$ to $e\_j$).