My attempt:
Let $\psi: \mathrm{End}_{R}(nM) \rightarrow M_n(\mathrm{End}_{R}(M)) $ given by $\phi \rightarrow (\phi_{ij})$ where $\phi_{ij} = \pi_j \circ \phi \circ i_i $, where $\pi_j$ is the projection and $i_i$ is the inclusion.
I need to show that $\psi$ is a ring isomorphism. So I need to show that
i) $\psi(\phi^1 + \phi^2) = \psi(\phi^1) + \psi(\phi^2)$
ii) $\psi(\phi^1 \circ \phi^2) = \psi(\phi^1) \psi(\phi^2)$
iii) $\psi$ is injective and surjective
We have that
i) $(\psi(\phi^1 + \phi^2) )_{ij} = \pi_j \circ (\phi^1 + \phi^2) \circ i_i=\pi_j \circ \phi^1 \circ i_i + \pi_j \circ \phi^2 \circ i_i = (\psi(\phi^1))_{ij} + (\psi(\phi^2))_{ij};\ \forall i,j.$
I am having trouble to prove ii) and iii)
for ii) I need to show that $C = \psi(\phi^1) \psi(\phi^2) = \psi(\phi^1 \circ \phi^2) = A$, i.é, $$C_{ij} = \sum_{k=1}^{n} = \phi^1_{ik}\phi^2_{kj} =\sum_{k=1}^{n}( \pi_k \circ \phi^1 \circ i_i \circ \pi_j \circ \phi^2 \circ i_k) = \pi_j \circ \phi^1 \circ \phi^2 \circ i_i = A_{ij};\ \forall i,j$$ but I don't know how to do that. I tried applying a generic $m \in M$, but I did not get the same thing on both sides.
for iii) I don't have nothing.
Thanks in advance.