linear algebra - Geometric proof of the Vandermonde determinant?

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 $Xcirc I = O$ ($circ$ 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.