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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 $(\Lambda^{top}Sym)\otimes (\Lambda^{top}Skew)^{-1}$.

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: In the real case, you can use conjugation by orthogonal matrices to reduce to the case where If $A$ is diagonal, and 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. The Over the real numbers, you can use conjugation by orthogonal matrices to reduce to diagonal case. The real version implies the general caseversion.
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 $\Lambda^{top}Sym/\Lambda^{top}Skew$. (\Lambda^{top}Sym)\otimes (\Lambda^{top}Skew)^{-1}$. 1 This is an interpretation of Terry Tao's answer (and BCnrd's comment). If$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$. Proof of FF: In the real case, you can use conjugation by orthogonal matrices to reduce to the case where$A$is diagonal, and 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. The real case implies the general case. If you want to turn this into a proof of the Vandermonde identity, then you have to find an independent reason for FF. 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/\Lambda^{top}Skew\$.