= Solution
For a complex square <matrix>, define its <determinant> by
$$
\det A=\sum_{\sigma\in S_n}\operatorname{sgn}(\sigma)\prod_{i=1}^na_{i,\sigma(i)}.
$$
The <cofactor> is $C_{ij}=(-1)^{i+j}\det A^{(ij)}$, where $A^{(ij)}$ deletes row $i$ and column $j$. The adjugate is the transposed <cofactor matrix>, $\operatorname{adj}(A)_{ij}=C_{ji}$.
The $(i,j)$ entry of $\operatorname{adj}(A)A$ is $\sum_kC_{ki}a_{kj}$. For $j=i$, expansion along column $i$ gives $\det A$. For $j\ne i$, this same sum is the <determinant> of the <matrix> obtained by replacing column $i$ by column $j$: deleting the replaced column leaves all its <cofactors> unchanged. The new <matrix> has two equal columns, so its <determinant> is zero. Thus
$$
\boxed{\operatorname{adj}(A)A=(\det A)I_n.}
$$
This proof does not assume invertibility and therefore also covers singular <matrices>.
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