adj(adj A) = (det A)^(n-2) A

Idea. Both sides are polynomial in the entries of $A$ (this is where $n\ge2$ matters: the exponent $n-2$ is non-negative). So it suffices to check the identity on the dense set $GL_n(\mathbb{R})$.

Step 1: facts about adj. Recall $A\operatorname{adj}A=\operatorname{adj}A\,A=(\det A)I$. So for invertible $A$, $\operatorname{adj}A=(\det A)A^{-1}$. Also, since each entry of $\operatorname{adj}X$ is a minor of size $n-1$,
$$\operatorname{adj}(cX)=c^{\,n-1}\operatorname{adj}X\qquad(c\in\mathbb{R}).$$

Step 2: invertible $A$. Then $\operatorname{adj}A=(\det A)A^{-1}$ is invertible with
$$\det(\operatorname{adj}A)=(\det A)^n\det(A^{-1})=(\det A)^{n-1}\ne0.$$
Apply the formula $\operatorname{adj}X=(\det X)X^{-1}$ to $X=\operatorname{adj}A$:
$$\operatorname{adj}(\operatorname{adj}A)=(\det A)^{n-1}\bigl((\det A)A^{-1}\bigr)^{-1}=(\det A)^{n-1}\cdot\frac{A}{\det A}=(\det A)^{n-2}A.$$
(Alternatively use homogeneity: $\operatorname{adj}\bigl((\det A)A^{-1}\bigr)=(\det A)^{n-1}\operatorname{adj}(A^{-1})$ and $\operatorname{adj}(A^{-1})=\det(A^{-1})A$.)

Step 3: density. The map $A\mapsto\operatorname{adj}(\operatorname{adj}A)-(\det A)^{n-2}A$ is polynomial in the entries (entries of $\operatorname{adj}$ are polynomials; the power $n-2\ge0$ is fine), hence continuous, and vanishes on $GL_n(\mathbb{R})$. As $GL_n(\mathbb{R})$ is dense in $M_n(\mathbb{R})$, it vanishes everywhere.

Check. For $n=2$, $\operatorname{adj}\begin{pmatrix}a&b\\c&d\end{pmatrix}=\begin{pmatrix}d&-b\\-c&a\end{pmatrix}$ and applying adj again returns the original matrix: this is $(\det A)^0A$.

Remark. For a singular matrix of rank $\le n-2$ both sides vanish when $n\ge3$: $\operatorname{adj}A=0$, so $\operatorname{adj}\operatorname{adj}A=0$ and $\det A=0$, so $(\det A)^{n-2}A=0$ (for $n\ge3$).

Lesson. Polynomial identities with a non-negative exponent extend from $GL_n$ to all of $M_n$ for free.