Show that if A is an nxn matrix then $\displaystyle det(adj A)=[det(A)]^{n-1}$
I'm not good with adjoints...first thing that comes to mind is $\displaystyle A(adj A)=(adj A)A=det(A)I_n$
Clearly you mean adjugate, and not adjoint. Use precisely the property you claimed. Namely, $\displaystyle \displaystyle \text{adj}(A)A=\det(A)I_n$. Taking the determinants of both sides gives $\displaystyle \det(\text{adj}(A))\det(A)=\det(\text{adj}(A)A)=\d et(\det(A)I_n)=\det^n(A)\det(I_n)=\det(A)^n$.
What you are calling the adjoint is what I call the adjugate. The adjoint of an endomorphism $\displaystyle A:V\to V$ (if it exists) on an inner product space $\displaystyle \left(V,\langle\cdot,\cdot\rangle\right)$ is the unique endomorphism $\displaystyle A^\ast:V\to V$ such that $\displaystyle \left\langle A(x),y\right\rangle=\left\langle x,A^\ast(y)\right\rangle$ for every $\displaystyle x,y\in V$. It has the property that if you fix an orthonormal ordered basis (assuming $\displaystyle \dim V<\infty$) $\displaystyle \mathcal{B}$ then $\displaystyle \left[A^\ast\right]_B=A\right]_B^\ast$ (where here $\displaystyle [\cdot]_B$ is the matrix representation with respect to that ordered basis, and $\displaystyle [A]_B^\ast=\left(\overline{[A]_B}\right)^{\top}$).
My linear book describes adjoint as the linear operator T* such that (T*(v), w) = (v, T(w)). * relates to linear operators in much the same way as * relates to matrices.
Of course, this isn't the first time mathematicians use the same word to describe multiple mathematical concepts.