The determinant of adjugate matrix Why does $\det(\text{adj}(A)) = 0$  if  $\det(A) = 0$? (without using the formula $\det(\text{adj}(A)) = \det(A)^{n-1}.)$
Use : 
$$\mathrm{adj}(A)\times A=A \times \mathrm{adj}(A)=\mathrm{det}(A)\mathrm{I}_{n}.$$

Let's assume that $\mathrm{det}(A)=0$. If $A=0$, then $\mathrm{adj}(A)=0$ as well and, obviously, $\mathrm{det}\big( \mathrm{adj}(A) \big)=0$. Otherwise, there exist $x \in \mathbb{R}^{n}$, $x \neq 0$ such that $Ax \neq 0$. It follows that $\mathrm{adj}(A) Ax=0$. As a consequence, $\mathrm{Span}(Ax) \subset \mathrm{ker} \; \mathrm{adj}(A)$ (which means that $\mathrm{ker} \; \mathrm{adj}(A) \neq \lbrace 0 \rbrace$). Therefore, $\mathrm{adj}(A)$ is not invertible and $\mathrm{det}\big( \mathrm{adj}(A) \big)=0$.
In that case, shall we use the following? $[adj(A)]^{-1} = adj(A^{-1})$ ?

Assume that $adj(A)$ is non-singular. Then inverse of adj(A) exists.

But we know $[adj(A)]^{-1}=adj(A^{-1})$ which is contradiction because it is given that $A$ is singular so $A^{-1}$ does not exist and hence adjoint of $A^{-1}$ also does not exist.

hence the proof.