We have $\pi(R) = \operatorname{End}_\mathbb{C}(V)$ is simple? 
Let $R$ be a $\mathbb{C}$-algebra and $(\pi, V )$ a finite dimensional simple $R$-module. Show that

(a) The ring $\pi(R) \subset \operatorname{End}_{\mathbb{C}}(V)$ is simple.
(b) We have $\pi(R) = \operatorname{End}_\mathbb{C}(V)$.
So I want to show that $\pi(R)$ is Artinian, and then show that its Jacobson radical is $0$, so then I can conclude it is semisimple. I'm not sure how to show it is Artinian, and where do I go after I show it is semisimple. Maybe work with the finite dimensionality of the endomorphism ring?
 A: Let us put $S=\pi(R)$ and $J$ its Jacobson radical. Note that $V$ is a simple $S$-module.
$S$ is semi-simple. Indeed, $S$ is artinian since its left ideals are complex subspaces of $S\subset\mathrm{End}_\Bbb{C}(V)$, which has finite dimension, and thus satisfy the descending chain condition. $S$ is also Jacobson semi-simple since by Nakayama's lemma and finite generation of $V$, $JV$ is a proper $S$-submodule of $V$ and therefore zero by simplicity of $V$ (*). It follows that $J=0$ and thus that $S$ is semi-simple.
(*) Alternatively, and more to the point, $JV=0$ by simplicity of $V$ as an $S$-module and by definition of the Jacobson radical as those elements in $S$ killing all simple $S$-modules.
$S$ is simple. By the Artin-Wedderburn theorem and the fact that there are no nontrivial finite dimensional complex division algebras we have $S\simeq\prod_{i=1}^rM_{n_i}(\Bbb{C})$. In fact, $r=1$ and $S\simeq M_m(\Bbb{C})$. Indeed, if there were two factors or more, the element $e$ of $S$ corresponding to $(I_{n_1},0,\dots,0)$ would define a nontrivial central idempotent of $S$, and thus two nontrivial $S$-submodules $\mathrm{Ker}(e)$ and $\mathrm{Im}(e)$, contradicting the simplicity of $V$ as an $S$-module.
$S=\mathrm{End}_\Bbb{C}(V)$. Recall that $M_m(\Bbb{C})$ has up to isomorphism a single simple module which is isomorphic to $\Bbb{C}^m$. We conclude that $V$ has dimension $\mathrm{dim}(V)=m$ and $S$ dimension $\mathrm{dim}(V)^2$. Therefore $S=\mathrm{End}_\Bbb{C}(V)$.
