I've been going through a proof of Wedderburn's theorem:
and I'm stuck on the very last part, where the author refers to example 2.1.4. (linked below). I don't understand what $D^n$ means, or why it would be isomorphic to $L$, or why this follows from the example 2.1.4. The chain of isomorphisms he refers to therefore makes absolutely no sense to me. If anyone could clear things up it would be much appreciated!

Example $2.1.4$ says in the last three lines that $M\simeq I_1^n=I_1\oplus \cdots \oplus I_1$. This is the isomorphism which you asked about. Hence we have $L\simeq D^n$, and analogously $L\simeq (D')^m$. If you want to have more details, see for example here: http://www.math.hawaii.edu/~lee/algebra/wedder.pdf.