Characteristic polynomial of $T:M_n(\mathbb{F}) \rightarrow M_n(\mathbb{F}) ,\ \ TX = AX \ \ (A\in M_n(\mathbb{F}))$

You have $(AE_{i,j})_{k,l}= 0$ if $l\neq j$ and $a_{ki}$ if $l=j$. Thus, if you consider the basis $(E_{11}, E_{21}, ..., E_{n1}, E_{12}, ... , E_{n2}, ... , E_{1n}, ... , E_{nn})$ of $M_n(\mathbb{F})$, then you're lead to computing the determinant of a bloc diagonal matrix of size $n^2\times n^2$, whose $n$ blocks are all equal to $A-XI_n$. This gives you that $\chi_T(X)=\chi_A(X)^n$.