If $\lambda_{1}$,$\lambda_{2}$,...,$\lambda_{n}$ are the eigenvalues of an $n\times n$ matrix $C$ then $\begin{matrix}\operatorname{trace}(C^k)=\lambda_{1}^k+\lambda_{2}^k+...+\lambda_{n}^k\end{matrix}$. Let $\lambda_{pi}$ be the $i$th eigenvalue of matrix $A_{p}$ where $1\leq p\leq m$ and $1\leq i\leq n$. Suppose that $A_{1}^k+A_{2}^k+...+A_{m}^k=O$ for every integer $k>0$ then $\begin{matrix}\operatorname{trace}(A_{1}^k)+\operatorname{trace}(A_{2}^k)+...+\operatorname{trace}(A_{m}^k)=0\end {matrix}$ and $\sum_{i=1}^{n}\lambda_{1i}^k+\sum_{i=1}^{n}\lambda_{2i}^k+...+\sum_{i=1}^{n}\lambda_{mi}^k=0$(*). Let $S$ be the set of all eigenvalues that are not equal to zero.Denote with $\mu_{j}$ the unique values of them and with $\nu_{j}$ the number of occurrences of that values where $1\leq j\leq q\leq mn$. Consider the homogeneous linear system of $q$ equations with uknowns $x_{1}$,$x_{2}$,...,$x_{q}$ $\begin{matrix} \mu_{1}x_{1}+\mu_{2}x_{2}+...+\mu_{q}x_{q}=0& & & \\ \mu_{1}^2x_{1}+\mu_{2}^2x_{2}+...+\mu_{q}^2x_{q}=0& & & \\ .& & & \\ .& & & \\ .& & & \\ \mu_{1}^qx_{1}+\mu_{2}^qx_{2}+...+\mu_{q}^qx_{q}=0& & & \end{matrix}$ From (*) we obtain that it has the non trivial solution $(\nu_{1},\nu_{2},...,\nu_{q})$ so the corresponding matrix $V$ for the uknowns which is Vandermonde matrix is singular.Consequently $\det(V)=\mu_{1}\mu_{2}...\mu_{q}\prod_{1\leq k<j\leq q}(\mu_{j}-\mu_{k})=0$ thus there exist $\mu_{j}=0$ or $\mu_{j}=\mu_{k}$ which contradicts the hypothesis for the set $S$. Hence all the eigenvalues are zeros therefore all matrices are nilpotent which is a contradiction.