Proof: $\mathrm{adj}(\mathrm{adj}(A)) = (\mathrm{det}(A))^{n-2} \cdot A$ for $A \in \mathbb{R}^{n\times n}$

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 $n\geq 2$.

Just replace $A$ with $Adj A$ in the identity

$$Adj A = \dfrac{|A| }{A}$$

And solve using other identities.