  Let $A$, $B$ be multidimensional matrices of boundary format respectively
$\prod_{i=0}^p(k_i+1)$, $\prod_{j=0}^q(l_j+1)$. Assume that $k_p=l_0$ so that
the convolution $A\ast B$ is defined. We prove that $Det (A\ast
B)=Det(A)^{\alpha}\cdot Det(B)^{\beta}$ where $\alpha= \frac {l_0!}{l_1!...
l_q!}$, $\beta=\frac {(k_0+1)!}{k_1! ... k_{p-1}!(k_p+1)!}$ and $Det$ is the
hyperdeterminant. When $A$, $B$ are square matrices this formula is the usual
Binet-Cauchy Theorem computing the determinant of the product $A\cdot B$. It
follows that $A \ast B$ is nondegenerate if and only if $A$ and $B$ are both
nondegenerate. We show by a counterexample that the assumption of boundary
format cannot be dropped.
