If is an eigenvalue of A with eigenvector x, then for all n, and hence . So the eigenvalues of are the exponentials of the eigenvalues of A; and the determinant in each case is the product of the eigenvalues.
Edit. Unfortunately, that answer was incomplete, because it does not deal with multiplicity. You need the matrices A and not only to have the same eigenvalues, but to have them with the same multiplicity. One way to do this is to see that the Jordan normal form of has the exponentials of the eigenvalues of A as its diagonal elements. In the answer I gave above, I was trying to avoid having to use the Jordan normal form, but maybe it's unavoidable.