Proof: $\mathrm{adj}(\mathrm{adj}(A)) = (\mathrm{det}(A))^{n-2} \cdot A$ for $A \in \mathbb{R}^{n\times n}$ I had my exam of linear algebra today and one of the questions was this one.
Given $n\ge2$ and $ A \in \mathbb{R}^{n \times n}$, prove that:
$$\mathrm{adj}(\mathrm{adj}(A)) = (\mathrm{det}(A))^{n-2} \cdot A.$$
Of course I was not able to prove this identity, otherwise I wouldn't post it here. But I'm still curious how one can prove this identity.
Could someone point me in the right direction?
 A: We use the identities
$$\tag 1\operatorname{adj}(A)\cdot A=\det A \cdot I_n$$
and $$\tag 2\operatorname{adj}(AB)=\operatorname{adj}(B)\cdot \operatorname{adj}(A).$$
We have by (1) 
$$\operatorname{adj}(\operatorname{adj}(A)\cdot A)=(\det A)^{n-1}\cdot I_n$$
and using (2)
$$\operatorname{adj}(A)\cdot \operatorname{adj}(\operatorname{adj}(A))=(\det A)^{n-1}I_n.$$
Multiplying by $A$ we get 
$$\det A \cdot I_n \cdot \operatorname{adj}(\operatorname{adj}(A))=(\det A)^{n-1}\cdot A.$$
If $\det A\neq 0$, we get the wanted equality, otherwise it's clear if $n\geq 2$.
A: Just replace $A$ with $Adj A$ in the identity
$$Adj A = \dfrac{|A| }{A}$$
And solve using other identities.
A: The equality holds over any commutative ring.
Short proof. It suffices to show that the equality holds for diagonal matrices, which is straightforward.
Slightly longer proof. Let 
$$
(a_{ij})_{i,j=1}^n
$$ 
be indeterminates. It is enough to check that the equality holds for the matrix 
$$
A\in M_n(\mathbb Q(a_{11},\dots,a_{nn}))
$$ 
whose $(i,j)$ entry is $a_{ij}$. But this clear since $A$ is semi-simple.
More details.
Why does it suffice to check the equality in this particular case?
Let $B$ be in $M_n(K)$, where $K$ is a commutative ring. The statement we must prove says that a certain matrix $F(B)$, depending on $B$, is zero. But each entry of $F(B)$ is a polynomial in the entries of $B$, and the coefficients of this polynomial are integers depending only on $n$.
Why is $A$ semi-simple?
Because the discriminant of its characteristic polynomial is nonzero.
