Let $M\in M_n(F)$ and define $\phi:M_n(F)\to M_n(F)$ by $\phi(X)=AX$ for all $X\in M_n(F)$. Prove that $\det(\phi)=\det(A)^n$.
I can prove it by considering the matrix representation with respect to the usual basis, which turns out to be a block diagonal form consisting of $n$ copies of $A$. Nevertheless, I'm looking for a clean (basis-free) approach to this problem.