Prove that if A is an n x n matrix, then det(adj A) = [det(A)]^(n-1).

I'm having a little trouble figuring this one out. I tried plugging in various equations to see if I could make the two sides equal to each other, but I'm getting stuck.

Right now I have that adj A = A^-1 * det(A)

If I plug that in for adj A then I get:

det(A^-1 * det(A)) = [det(A)]^(n-1), but I'm getting stuck on the right side of the equation. I don't see how the (n-1) part comes in.

I tried plugging in equations for det(A)

I tried det(A) = adj(A) * A * In^-1 but I deaded ended with that equation too.

Any help about how to get started would be appreciated! Thanks!