Adjugate identities by density

Reminder. $\operatorname{adj}A$ is the transpose of the cofactor matrix; its entries are polynomials (of degree $n-1$) in the entries of $A$, and $A\operatorname{adj}A=\operatorname{adj}A\cdot A=(\det A)I$. On $GL_n$ this gives $\operatorname{adj}A=\det(A)A^{-1}$.

First identity on $GL_n\times GL_n$. For invertible $A,B$,
$$\operatorname{adj}(AB)=\det(AB)(AB)^{-1}=\det A\det B\,B^{-1}A^{-1}=\bigl(\det B\,B^{-1}\bigr)\bigl(\det A\,A^{-1}\bigr)=\operatorname{adj}B\operatorname{adj}A.$$

Second identity on $GL_n$. The matrix $\operatorname{adj}A=\det A\cdot A^{-1}$ is a scalar multiple $c=\det A$ of $A^{-1}$, and $\det(cM)=c^n\det M$, so
$$\det\operatorname{adj}A=(\det A)^n\det(A^{-1})=(\det A)^n(\det A)^{-1}=(\det A)^{n-1}.$$

Extension. Each side of each identity is a polynomial in the entries of $A$ (and $B$). A polynomial identity valid on the dense set $GL_n\times GL_n$ (dense in $M_n\times M_n$) holds everywhere, by continuity.

Check. $n=2$: $\operatorname{adj}\begin{pmatrix}a&b\\c&d\end{pmatrix}=\begin{pmatrix}d&-b\\-c&a\end{pmatrix}$ and $\det\operatorname{adj}A=ad-bc=(\det A)^1$.

Remark. For $A$ of rank $n-1$, $\operatorname{adj}A$ has rank $1$, and for rank $\le n-2$ it is $0$; the identity $\det\operatorname{adj}A=(\det A)^{n-1}=0$ is consistent when $n\ge2$. Lesson. Identities with $A^{-1}$ become identities with $\operatorname{adj}A$ after multiplying by $\det A$, which makes them polynomial.