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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), a combinatorial proof by Art Benjamin and Gregory Dresden (here), and another (slightly less) combinatorial proof by Jennifer Quinn (here, 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.

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The question is really nice but it's probably hard to interpret purely geometrically the powers of $x_j$, the algebraic "addition" in Charles's argument seems to be unavoidable. It's probably a good reason to understand the product $\prod_{1\le i<j\le n}(x_i-x_j)$ from a geometric point of view. It's square, the discriminant, possesses some "hidden" geometry for small $n$ (in the theory of elliptic curves, for example) but probably not the product itself... – Wadim Zudilin Jun 24 '10 at 10:47
Obviously I'm OK with some algebraic manipulation. That said, the type of proof I had in mind might be something like: take a rectangular prism containing the parallelepiped in question and subtract out the volume of some leftover simplices. Or find some other polytope with the same volume and prove synthetically that their volumes match. – Daniel Litt Jun 24 '10 at 12:56
Thanks, Daniel, for clarification: you'd like to have some "measure" interpretation of the involved polynomials rather than view them as defining algebraic curves/surfaces. An interesting restriction! I try to find out whether the product of $x_i-x_j$ is a volume of some meaningful parallelepiped... – Wadim Zudilin Jun 24 '10 at 13:20
Well, my motivation is that the row-reduction proof makes the geometry really obvious---essentially, row reduction finds a volume-preserving linear transformation that turns the parallelepiped defined by the Vandermonde matrix into a rectangular prism. I'd just like to have a more synthetic proof. – Daniel Litt Jun 24 '10 at 13:42

11 Answers 11

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).

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This is very nice as well! I would still not say it is geometric however--perhaps the criterion I have in mind is "something one can draw." (Sorry for my lack of generosity :-/.) Also, it seems to me that the last step does sort of boil down to row-reduction, no? Regardless, a beautiful argument of a different flavor from those we've seen thus far! – Daniel Litt Jun 24 '10 at 0:37
yes, I knew from the beginning that you wanted an n-dimensional volume... but what embarasses me is that the product on the RHS has n(n+1)/2 factors, and I can't imagine what to do with them... – Pietro Majer Jun 24 '10 at 0:43
This argument can be given geometrically (with a picture, and without calculation), but it works only for proving that $V$ is a function of the differences $a_i - a_j$, i.e., is translation invariant. Here $f(t)$ is constant because it is the Wronskian of the differential equation $D^n = 0$. Seen as a linear ODE on an $n$ dimensional phase space, this has trace 0 and therefore is a volume-preserving flow. That's quite geometric, but Vandermonde's formula asks about the volume of a specific simplex defined by the $a_i$, and not the evolution of that simplex over time. – T.. Jun 25 '10 at 20:07
T: Could you possibly flesh this out in an answer? – Daniel Litt Jun 25 '10 at 21:50
@T: of course it's a Wronskian, and I was wondering whether to remark it or not. Anyway translation invariance can be deduced after just a derivate, and that's it. Translation invariance implies easily the product formula, because it allows to perform the computation inductively on n. – Pietro Majer Jun 25 '10 at 22:34

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.

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This is essentially the algebra-geometric argument I give in the question, although a bit more geometric than mine. While this is a wonderful way to look at it, I'm more looking for a proof via Euclidean geometry. Although any proofs are welcome--this argument is great! – Daniel Litt Jun 23 '10 at 17:15
Ahh, I admit, I skimmed the question and the argument there didn't immediately ring any bells, so I figured I'd post this one. I don't know any Euclidean geometry proofs. – Charles Siegel Jun 23 '10 at 17:53
I'd also like to point out that there's no reason to think that the coefficient does not depend on $n$ a priori; this subtlety can be dealt with as follows. It is easy to narrow down the coefficient to $\pm 1$ (essentially by inspection); then note that in general if $x_1 < x_2 < \cdots < x_n$, the determinant is positive. – Daniel Litt Jun 23 '10 at 18:52
For the proof of the lemma, I think it is cleaner to say that $n$ $\textrm{distinct}$ points on the affine rational normal curve $t \to (1, t,t^2,\ldots,t^{n-1})$ are not contained in any hyperplane by the fundamental theorem of algebra (who knows what goes into the basics of "the notion of degree"?). Also, note that the lemma claims the more complicated direction "only if" in your second paragraph, whereas what's needed for the Vandermonde formula is the easier "if" direction, as in Daniel's sketch. – Victor Protsak Jun 25 '10 at 7:54
How do you prove that $n$ points on the curve don't satisfy a linear relation, without at least showing Vandermonde determinant is nonzero (and how do you show it's nonzero without having the full formula with the product of $(x_i - x_j)$'s )?? For example, the same curve with $t^{n-1}$ replaced by $t^n$ does have linear relations between points on the hyperplane $x_1 + x_2 + \dots + x_n = 0$. – T.. Jun 25 '10 at 19:39

This is an interpretation of Terry Tao's answer (and BCnrd's comment).

If $A$ is $n\times n$ symmetric then $ad_A:X\mapsto AX-XA$ maps symmetric matrices to skew matrices. Generically this is surjective, and generically its kernel has the first $n$ powers of $A$ as a basis. Choosing bases for the symmetric matrices and for the skew matrices (independent of $A$!), you then have a determinant to be computed, which appears to depend on the generic matrix $A$. However, we have -

Funny Fact: This is independent of $A$.

Proof of FF: If $A$ is diagonal, then a computation using the first bases you think of shows that this determinant is the quotient of a Vandermonde determinant and the usual expression for the same. Over the real numbers, you can use conjugation by orthogonal matrices to reduce to diagonal case. The real version implies the general version.

If you want to turn this into a proof of the Vandermonde identity, then you have to find an independent reason for FF. I do not have one to offer.

A cool restatement of FF is:

Although the basis $1, A, \dots , A^{n-1}$ for $ker(ad_A)$ is (of course) dependent on $A$, the generator which it gives you for the top exterior power of this vector space does not depend on $A$. Here I am using the short exact sequence $0\to ker(ad_A) \to Sym\to Skew\to 0$ to identify the $1$-dimensional vector spaces $\Lambda^n ker(ad_A)$ and $(\Lambda^{top}Sym)\otimes (\Lambda^{top}Skew)^{-1}$.

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I think this can't quite be true as stated. $ad_A: Sym \to Skew$ is not an isomorphism (though as you point out it is generically surjective). It only becomes one when viewed as a map $Sym/\ker(ad_A)\to Skew$ or, as Terry Tao does, when restricting to the orthogonal complement of $\ker(ad_A)$. So the determinant doesn't make sense except out of these spaces. But how do you pick bases of these spaces independently of $A$? They're defined using $A$. It seems to me you want to compute the determinant of an $n(n-1)/2\times n(n+1)/2$ matrix. Am I not seeing something? – Daniel Litt Jun 25 '10 at 15:44
When you have a short exact sequence of finite-dimensional vector spaces 0->U->V->W->0, then a choices of ordered basis for all three spaces gives you a well-defined number, the determinant of a matrix that compares the chosen basis of V with another basis consisting of the image of the basis of U and any lifting of the basis of W. The indeterminacy in the lifting has no effect. A fancier way to say this that such a sequence gives you an isomorphism between two 1-dimensional vector spaces: the top exterior power of V and the tensor product of the top exterior powers of U and of W. – Tom Goodwillie Jun 25 '10 at 17:07
Carry this out for the sequence 0->ker(ad_A)->Sym->Skew->0 when A is diagonal (with distinct eigenvalues), using your favorite bases for skew and sym and the powers-of-A basis for the kernel, and you will run into a block sum of (1) an nxn Vandermonde matrix and (2) an NxN diagonal matrix where N=dim(Skew). The latter is (up to signs) the inverse of a matrix whose diagonal entries are the x_i-x_j – Tom Goodwillie Jun 25 '10 at 17:08
Ah I see. I upvoted your comment; your answer does indeed make sense (as I'm sure you knew). – Daniel Litt Jun 25 '10 at 17:34
This argument seems closely related to the fact that $\det (f_i(x_j))$ is divisible by $\Pi (x_i - x_j)$, for arbitrary polynomials (or functions) $f_i$. Here the external proof of the Fun Fact comes from the degree argument used to prove the Vandermonde identity; the determinant divided by the product is a polynomial of degree zero. – T.. Jun 25 '10 at 20:36

I have what looks like the first half of an answer, but for some strange reason I can't see how to finish the job.

Let $A$ be an $n \times n$ real matrix with eigenvalues $x_1,...,x_n$. On the one hand, the adjoint operator $ad(A): B \mapsto [A,B]$, acting from the space $C(A)^\perp$ of symmetric matrices orthogonal to centraliser of A to the space of skew-symmetric matrices, has determinant $\prod_{1 \leq i < j \leq n} (x_i - x_j)$ (as can be seen by diagonalising $A$, and after fixing some sign conventions).

On the other hand, generically the centraliser $C(A)$ is an $n$-dimensional space spanned by $1, A, \ldots, A^{n-1}$, and the determinant of this basis is $\det(x_j^{i-1})_{1 \leq i < j \leq n}$ (again up to some normalisations).

So presumably there must be some special property of the basis $1,A,\ldots,A^{n-1}$ of the kernel of the adjoint operator $ad(A)$ which would connect the two quantities and finish the job, but I can't see it yet, though it looks very close; I have a vague feeling one wants to work somehow in the non-commutative polynomial ring $M_n({\bf R})(X)$ formed by adjoining an non-commutative indeterminate X to the matrix ring $M_n({\bf R})$, and then specialise X to A, but my algebra is not good enough to push this through immediately. The fact that $\det(T^*) = \det(T)$ for any linear transformation T also seems relevant somehow.

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Terry, maybe instead of bringing in an orthogonal complement it is "more natural" (and may simplify some of the "normalization" calculations) to work with the quotient space of symmetric matrices modulo the centralizer of A. – BCnrd Jun 24 '10 at 5:56
Ideally we would have some more intrinsic characterization of $\det[1, A, A^2, ..., A^{n-1}]$. There's clearly something here, but I am (unsurprisingly) not having any luck either. – Daniel Litt Jun 24 '10 at 19:04
This looks like a circular argument because the second paragraph ("generically...") includes a restatement of Vandermonde formula in less precise language. Re "I have a vague feeling": the standard trick is to consider the ring $M_n(R)[\lambda]$ of polynomials in $\lambda$ over the $n\times n$ matrices over $R=k[a_{ij}]_{1\leq i,j\leq n}$ polynomials in the indeterminates $a_{ij}$ (the entries of a generic matrix $A$) over the base ring $k$, prove some identity there, then specialize $\lambda\to A.$ – Victor Protsak Jun 25 '10 at 8:19
@Victor: I don't think it's circular---you just need to know that the determinant is generically non-vanishing, which is the same as it not vanishing identically. But this is clear from a variety of considerations. – Daniel Litt Jun 25 '10 at 13:03
re: "not vanishing identically", how do you find a point (or points) where the determinant doesn't vanish, without assuming the Vandermonde formula? Doing this deterministically for nonzero multivariable polynomials is a nontrivial (and in some respects unsolved) problem, so you have to use specific structure of this determinant, and even so I don't see how to circumvent the use of the formula. – T.. Jun 25 '10 at 20:24

Let $F$ be a field, and $\mathbb{K}$ be the field of fractions of the polynomial ring $R = F[x_{1},x_{2},\ldots x_{n}].$

Take $n$ mutually projections from $\mathbb{K}^{n}$ onto $1$-dimensional spaces, $\{E_{i}: 1 \leq i \leq n \}.$ If you like, let $E_{i}$ be projection onto the $1$-dimensional space of row vectors in $\mathbb{K}^{n}$ with $0$ in place $j$ for $j \neq i.$ Consider the linear transformation $X = \sum_{i=1}^{n} x_{i}E_{i}.$ I claim that $\{ I,X,X^{2},\ldots , X^{n-1} \}$ is linearly independent and has the same linear span as $\{E_{1},E_{2},\ldots, E_{n} \}.$ Clearly the former span is contained in the span of the idempotents. For each $i,$ let $P_{i} = \prod_{j \neq i} \frac{X-x_{j}I}{x_{i}-x_{j}}.$ Now $P_{i} \neq 0$ for any $i$ because $XE_{i} = x_{i}E_{i}$ for each $i.$ In fact, we have $P_{i}E_{j} = 0$ for $j \neq i,$ and $P_{i}E_{i} = E_{i}.$ Since $P_{i}$ is in the linear span of the $E_{k}s,$ we must have $P_{i} = E_{i}$ for each $i.$ Hence each $E_{i}$ is in the linear span of $\{I,X,\ldots, X^{n-1}\}.$ This implies that the rows of the Vandermonde matrix are linearly independent, and that its determinant divides $\prod_{i < j}(x_{i}- x_{j})^{2}.$ In fact, we see that $E_{k} \prod_{i < j}(x_{i}-x_{j}) $ is an $R$-combination of $\{I,X,\ldots ,X^{n-1} \}$ for each $k,$ which implies that the determinant of the Van der Monde matrix divides $\prod_{i < j} (x_{i}-x_{j})$ in $R.$ Consideration of the coefficient of $X^{n-1}$ in each $E_{k}$ shows that this product must also divide the determinant in $R.$

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Ira Gessel used transitive tournaments in graphs to prove Vandermonde’s determinant identity: This proof certainly has some geometric flavour although not in the initial sense of the question.

P.S. The historical process that led to the worldwide adoption of the denomination "Vandermonde determinant" is studied in (A case of mathematical eponymy: the Vandermonde determinant, by Bernard Ycart).

Interestingly, $3\times 3$ Vandermonde determinant turns out to be (up to the square root) the correct physical variable of the odderon due to modular invariance of the odderon: (The Odderon and Invariants of Elliptic Curves, by Romuald A. Janik).

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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?

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It sounds like – Wadim Zudilin Jun 25 '10 at 4:41
I think this is exactly the same as row reduction if you write it out, no? That said, it's a great way of doing the computation. – Daniel Litt Jun 25 '10 at 13:05
@Daniel: yes, if you look at the row reduction argument given in your link, it's actually what you get by doing only the column operations, which is how I found it (since I wanted to see what kind of affine transformation this produced). The matrix can be factored into block unipotent matrices which have $n−1−j$ copies of $−x_j$ just above the diagonal in the lower right corner, for $j=1,\ldots,n−1$. – Ian Agol Jun 25 '10 at 16:35
One can phrase this a little bit more geometrically as follows. The $n$-dimensional space $V$ of polynomials of degree less than $n$ has two bases: the standard basis $1,x,\ldots,x^{n-1}$, and Agol's basis $1, (x-x_1), (x-x_1)(x-x_2),\ldots,(x-x_1)\ldots(x-x_n)$. It also has the map $\Phi: P \mapsto (P(x_1),\ldots,P(x_n))$ from $V$ to $R^n$. The Vandermonde det is det of $\Phi$ with respect to the standard basis, while the product $\prod_{1 \leq i < j \leq n} (x_i-x_j)$ is the det of $\Phi$ wrt Agol's basis (where it becomes triangular). But the two bases are linked by a unipotent map. – Terry Tao Oct 31 '10 at 17:46
Similarly, with $e_k(x_1,...,x_n)$ the elementary symmetric polynomials, $de_1 \wedge de_2 \; ...\wedge de_n= |V_n|\; dx_1 \wedge dx_2 \; ...\wedge dx_n.$ – Tom Copeland May 19 '14 at 1:23

A professor of mine brought this question up during a seminar the other day, and I offered a "peudo-geometric" answer, interpreting the Gaussian elimination proof in terms of volume preserving shear maps $\tau_{\lambda} \in End(R^n)$ of the form $(x_1,x_2,\dots,x_n) \mapsto (x_1-\lambda,x_2-\lambda x_1,\dots,x_n-\lambda x_{n-1})$ for $\lambda \in R$ scalar. Essentially, this argument boiled down to the fact that Gaussian row-reduction operations are "well-behaved" with respect to volume. I wonder if this cuts it?

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This is really the geometric interpretation I allude to in the last 2 sentences of the question. But I do think this is an illuminating interpretation of what row reduction "really" is. – Daniel Litt Dec 2 '10 at 18:40

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 $e_j$).

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I would have posted this as a comment, but it's too long for comment, so I post it here. Here is my version of a sketch a geometric proof for nonsingularity of it when $x_i$'s are distinct, but I don't think that I can improve it to find the determinant:

Two manifolds $P$ and $S$ are called to intersect transversally at a point $A$, if the tangent spaces of $P$ and $S$ together span the whole ambient space. Let $\lambda_1,\ldots, \lambda_n$ be distinct real numbers, and let $A$ be the diagonal matrix $A=\rm{diag}(\lambda_1,\ldots,\lambda_n)$, and let $S$ be the set of all matrices with the same spectrum as $A$. In a small neighborhood around $A$, $S$ becomes a manifold. Let $P$ be the manifold of all the diagonal matrices of size $n$. The tangent space of $S$ and $P$ at $A$ can be computed and shown that the intersection is transversal.

On the other hand, define a function $f$ that maps any diagonal matrix $B$ (with $x_i$'s on its main diagonal) to $(\frac{\rm{tr}B}{1},\frac{\rm{tr}B^2}{2},\dots,\frac{\rm{tr}B^n}{n})$. It can be seen that the Jacobian of $f$ evaluated at $A=\rm{diag}(\lambda_1,\ldots,\lambda_n)$ is the Vandermonde matrix, which is nonsingluar if and only if $\lambda_i$'s are distinct.

Putting the two pieces above together, and with a little bit of discussion, one can show that $\rm{Jac}(f) \big|_A$ being nonsingular is equivalent to having $P$ and $S$ intersect transversally at $A$.

One can see the relations of the above approach to the Terry Tao's answer by noting that the tangent space to $S$ at $A$ is the set $\{[B,A] : B \text{ is a skew-symmetric matrix}\}$.

One relation to the powers of $A$ comes form a way of showing the above Jacobian matrix is nonsingular. Note that $$\rm{Jac}(f)\big|_A = \left[ \begin{array}{} I_{11} & I_{22} & \cdots & I_{nn}\\ A_{11} & A_{22} & \cdots & A_{nn} \\ \vdots & \vdots & \ddots & \vdots\\ A^{n-1}_{11} & A^{n-1}_{22} & \cdots & A^{n-1}_{nn}\end{array} \right]$$ In order to show the nonsingularity above assume that $\left[\begin{array}{} \alpha_1, \ldots, \alpha_n \end{array} \right] \rm{Jac}(f)\big|_A = 0$. This means if you consider the polynomial $p(x) = \sum_{i=1}^{n} \alpha_i x^{i-1}$ and let $X = p(A)$, you want to show if $X\circ I = O$ ($\circ$ is the Schur product) then $X=O$, but that is easy to show, since $A$ is diagonal, hence $p(A)$ is, and so $p(x)$ has $n$ distinct roots, but $\rm{deg}(p(x))=n-1$, thus $p(x)$ is the zero polynomial.

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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 $\prod_{i<j} (j-i)=0!1!\dots (n-1)!$.

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