Why $\det(L_A)=[\det(A)]^n$? $A$ is an $n\times n$ matrix and $L_A$ is the left-multiplication operator on the $n\times n$ matrices.
I have seen this question and the given answer. But I could not understand the answer. I have commented on the answer, but did not get any reply for more details. 

So, could any one please give me some more details.

 A: We're viewing left multiplication of $A$ as a linear operator on the $n^2$-dimensional vector space of $n\times n$ matrices.
One convenient basis for the $n^2$-dimensional space consists of all matrices that have $0$ in every entry except $1$ in one of them. $L_A$ maps each of these basis elements to a matrix that contains one column from $A$ and zeroes in all the other columns. If we write down the matrix of $L_A$ with respect to this basis (assuming we have ordered the basis in column-major order), we get this $n^2\times n^2$ block matrix:
$$ \begin{pmatrix} A & 0 & \cdots & 0 \\
 0 & A &\cdots & 0 \\
\vdots &\vdots & \ddots & \vdots \\
0 & 0 &\cdots & A \end{pmatrix} $$
It is an elementary fact that the determinant of a block diagonal matrix is the product of the determinants of the blocks on the diagonal -- in this case, $(\det A)^n$.
(This can be seen by considering the full expansion of the determinant, for example, or by writing each of the blocks as a product of elementary matrices).
