If the matrix rings are isomorphic, then the scalars rings are isomorphic I'm solving exercises of noncommutative ring theory and I have find across the following problem.

If $D$ and $D'$ are division rings and $M_m(D)\simeq M_n(D')$, show that $D\simeq D'$ and $m=n$.

I have already tried to attack in various ways and I understand that this exercise says that, by the Wedderburn-Artin theorem, a simple Artinian ring $R$ is a ring of matrices over a division ring $D$ unique up to isomorphism.
Any suggestion is appreciated. Thank you.
 A: (I just expanded on the answer of Torsten Schoeneberg in the comments above)
We have the following ring isomorphisms for any division ring $D$:
(i) $\Gamma: D^{o}\rightarrow \text{End}_D(D)=\{D\text{-module endomorphisms of }D\}\\d\mapsto \theta_d\quad \text{s.t.}\quad \theta_d(x)=xd$
(ii) $\Gamma: D^{o}\rightarrow \text{End}_{M_n(D)}(D^n)=\{M_n(D)\text{-module endomorphisms of }D^n\}\\d\mapsto \theta_d\quad \text{s.t.}\quad \theta_d(x_1,\ldots,x_n)=(x_1d,\ldots,x_nd)$
(iii) $\Gamma: M_n(D)^o\rightarrow M_n(D^o)\\A\mapsto A^T$
(iv) $\Gamma: M_n(D^o)\rightarrow \text{End}_D(D^n)\\A\mapsto \theta_A\quad \text{s.t.}\quad \theta_A(x_1,\ldots,x_n)=(x_1,\ldots,x_n).A$
Using the second ring isomorphism, we have the following argument:
Given two division rings $D_i$, $i=1,2$,
$D_i^n$ is the unique simple $M_n(D_i)$-module up to isomorphism, so, since $M_n(D_i)$ are isomorphic as rings by hypothesis, $D_i^n$ are isomorphic as $M_n(D_i)$-modules, so $D_1^o\cong \text{End}_{M_n(D_1)}(D_1^n)\cong \text{End}_{M_n(D_2)}(D_2^n)\cong D_2^o$, so $D_1\cong D_2$.
