Problem
(i) Let $R,T$ be division rings and $m,n \in \mathbb N$, then $M_n(R \times T) \cong M_n(R) \times M_n(T)$ and $M_m(M_n(R)) \cong M_{mn}(R)$.
(ii) If $R$ is a semisimple ring and $n \in \mathbb N$ then $M_n(R)$ is semisimple.
Using (i) I could show (ii):
Suppose $R$ is a semisimple ring, then by Wedderburn theorem there are $D_1,...,D_r$ division rings and $n_1,...,n_r$ such that $R \cong M_{n_1}(D_1) \times ... \times M_{n_r}(D_r)$. So $M_n(R) \cong M_n(M_{n_1}(D_1) \times ... \times M_{n_r}) \cong M_{nn_1}(D_1) \times ... \times M_{nn_r}(D_r)$, and again, by Wedderburn this implies $M_n(R)$ is semisimple.
I would appreciate if someone could help me to show the properties in (i).
I guess you don't really have a problem with the $M_n(R\times T)\cong M_n(R)\times M_n(T)$ (or if you do, you'll probably see the solution soon for yourself.)
Hints:
If you believe that matrices can be multiplied "in blocks", then the isomorphism $M_m(M_n(R)) \cong M_{mn}(R)$ is just a reflection of that. Given an element $((a_{i,j}))_{x,y}\in M_m(M_n(R))$, you can make it into a block matrix in $M_{mn}(R)$ by sending $((A)_{i,j})_{x,y}\mapsto (A)_{xn+i,yn+j}$. However, this seems rather complex to verify if you have doubts about "block multiplication" working.
Alternatively, you could try to reason abstractly by using the fact that $M_n(R)$ is isomorphic to the ring of $R$-linear tranformations of $R^n$, $End(R^n)$. You would be arguing that for each $End(R^n)$-linear transformation of $End(R^n)^m$, you can produce an $R$ linear homomorphism of $R^{mn}$ such that the assignment is a ring homomorphism. The obvious candidate for a mapping is to let the elements of $End(R^n)$ operate on length $n$ blocks of an element of $R^{mn}$.