Show that if $A$ and $B$ are invertible $n\times n$ matrices then
$$\det\left((A^T)^7B^{-15}A^{29}(B^T)^{11}A^{-36}B^4\right)=1.$$
I'm guessing the expression inside the brackets simplifies to the identity matrix, but I'm having trouble simplifying this expression at all. How do you do this?
\begin{align} & \det((A^T)^7 \times B^{-15} \times A^{29} \times (B^T)^{11} \times A^{-36} \times B^4) \\ &= \det((A^T)^7) \det(B^{-15}) \det(A^{29}) \det((B^T)^{11}) \det((A)^{-36}) \det(B^4) \tag{$\det(AB) = \det(A)\det(B)$}\\ &= \det(A^T)^7 \det(B)^{-15} \det(A)^{29} \det(B^T)^{11} \det(A)^{-36} \det(B)^4 \\ &= \det(A)^7 \det(B)^{-15} \det(A)^{29} \det(B)^{11} \det(A)^{-36} \det(B)^4 \tag{$\det(A^T) = \det(A)$}\\ &= \det(A)^{7+29-36} \det(B)^{-15+11+4} \\ &= 1 \end{align}