Said more specifically, suppose $A,B\in M_n(K)$, $K$ a commutative ring. An $r$-th principal minor of a square matrix is the determinant $$\det\begin{bmatrix}a_{k_1k_1} & \cdots & a_{k_1k_r}\\ \vdots & \ddots & \vdots \\ a_{k_rk_1} & \cdots & a_{k_rk_r}\end{bmatrix}$$ where $1\le k_1<\cdots<k_r\le n$, $A$ is an $n\times n, n>r$ square matrix. Prove that the sum of all of $AB$'s $r$-th principal minors is equal to that of those of $BA$'s.
Hint: use Cauchy-Binet formula.
I simply have no idea what use to be made of the hint. I just can't represent the terms in the way Cauchy-Binet formula does, anyway for me it seems impossible to construct any of $P_r(AB)$ from sub-blocks of $A,B$.
Any help?
Edit: I'm sorry I made a mistake. It should be principal minor, no leading.
You may begin with $P_r(AB)=\det(A_{[r],[n]}B_{[n],[r]})$ and apply Cauchy-Binet formula to decompose it into a sum of products of determinants.