For a finite-dimensional algebra, the Jacobson radical is nilpotent and the quotient by it is semisimple

The left regular module ${}_{A}A$ is finite-dimensional, so [L1] makes the descending chain $A\supseteq J\supseteq J^2\supseteq \cdots$ stabilize: choose $n$ with $J^n=J^{n+1}$. If $J^n\mathrm{\ne }0$, [L2] gives a maximal submodule $N$ of the finite-dimensional left $A$-module $J^n$, so $S=J^n\mathrm{/}N$ is simple by [L3]. For every nonzero $s\in S$, the map $A\to S$, $a\mapsto as$, is onto; its kernel is a maximal left ideal, hence contains $J$ by [F1]. Thus $Js=0$ for every $s$, so $JS=0$ and $J^{n+1}=J{J}^n\subseteq N$, contradicting $J^{n+1}=J^n$. Therefore $J^n=0$, so $J$ is nilpotent.

Maximal left ideals of $A\mathrm{/}J$ are exactly the quotients $L\mathrm{/}J$ with $L$ a maximal left ideal of $A$ containing $J$, so their intersection is $0$ by [F1]. Let $M={}_{A\mathrm{/}J}(A\mathrm{/}J)$ be the left regular module. It is finite-dimensional. We prove by induction on vector-space dimension that any finite-dimensional module whose maximal submodules intersect trivially is semisimple. If $M=0$ there is nothing to prove. Otherwise choose a nonzero submodule $S\le M$ of least positive dimension, so $S$ is simple. Since the intersection of maximal submodules is $0$, some maximal submodule $N$ does not contain $S$. Then $S\cap N=0$, and maximality makes $(S+N)\mathrm{/}N$ a nonzero submodule of the simple quotient $M\mathrm{/}N$, so $M=S\oplus N$. For any nonzero $x\in N$, choose a maximal submodule $L$ of $M$ avoiding $x$. The image of $N$ in the simple module $M\mathrm{/}L$ is nonzero and hence all of $M\mathrm{/}L$, so $N\cap L$ is a maximal submodule of $N$ avoiding $x$. Thus the maximal submodules of $N$ also intersect trivially. Its dimension is smaller than that of $M$, so induction makes $N$ semisimple, hence so is $M$.

By step 1.2, the left regular module of $A\mathrm{/}J$ is semisimple. Thus [L4] makes the quotient algebra $A\mathrm{/}J$ semisimple. Together with step 1.1, this proves the theorem.