The determinant of adjugate matrix I have the following proof that I would like to be walked through because I'm not intuitively seeing what to do:
If $A$ is $n\times n$, prove $\det\left(\operatorname{adj}(A)\right) = \det(A)^{n-1}$.
I know the property of $A\operatorname{adj}(A) = \det(A)I$ is important but I don't know how to apply it to get an answer. Any help is much appreciated.
Thanks,

We have the relation
$$A \operatorname{adj}(A)=\det(A)I $$
Now take determinant on both the sides we get,
$$\det(A)\det(\operatorname{adj}(A))=\det(\det(A)I) \tag1$$
use the relation that, for a matrix $A$ that is $n \times n$,
$$\det(kA)=k^n (\det(A))$$
where k is a numerical constant. Hence we have
$$\det(\det(A)I)=\det(A)^n \det (I)=\det(A)^n \tag2$$
and from equations $(1)$ and $(2)$ it follows that
$$\det(A)\det(\operatorname{adj}(A))=\det(A)^n ,$$
which implies
$$\det(\operatorname{adj}(A)) = \det(A)^{n-1} .$$
