How to prove that if $\det(A)=0$ then $\det(\operatorname{adj}(A))=0$?

Let $B$ denote the adjugate matrix of $A$. Suppose for the sake of contradiction that $\det(B) \neq 0$. Then $B$ is invertible. Since the equation $AB =(\det{A})I = 0$ is true, we have then $$AB\vec{v} = \vec{0}\ \ \forall \vec{v}$$ which implies $A$ is the zero matrix. But then the adjugate of the $0$ matrix is clearly $0$ itself which contradicts the fact that $B$ was invertible.

Let's write $A'$ for the adjoint of $A$. $AA'=(\det A)I$, so $\det A\det A'=(\det A)^n$ (where $A$ is an $n\times n$ matrix). If $\det A\ne0$, this yields $\det A'=(\det A)^{n-1}$. By continuity, this last equation is true even when $\det A=0$, and you're done.