Prove this determinant identity combinatorially

I am very surprised that no solution was given yet.

Denote by $\mathrm{Com}(A)$ the comatrix of $A$. The determinant to be computed is the determinant of the product $D \times \mathrm{Com}(A) \times D$, where $D$ is the diagonal matrix with diagonal entries $((-1)^i)_{1 \le i \le n}$. Since $D^2=I_n$, this matrix is similar to $\mathrm{Com}(A)$, so it as the same determinant as $\mathrm{Com}(A)$.

Since $A \times \mathrm{Com}(A)^\top = (\det A) I_n$, we have $\det(A) \times \det(\mathrm{Com}(A)) = \det((\det A) I_n) = (\det A)^n$.

If $A$ is invertible, we derive $\det(\mathrm{Com}(A)) = (\det A)^{n-1}$.

If $A$ is not invertible and not null, then $\mathrm{Com}(A)$ cannot be invertible since $A \times \mathrm{Com}(A)^\top$ is null, so $\det(\mathrm{Com}(A))$ is null.

If $A$ is null and $n \ge 2$, then $\mathrm{Com}(A)$ is null.

If $A$ is null and $n=1$, then $\mathrm{Com}(A)$ is the $1 \times 1$ matrix with unique entry equal to $1$.

In all cases, we derive $\det(\mathrm{Com}(A)) = (\det A)^{n-1}$.