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] gives finite length. Therefore the descending chain $A\supseteq J\supseteq J^2\supseteq \cdots$ stabilizes: choose $n$ with $J^n=J^{n+1}$. If $J^n\mathrm{\ne }0$, [L2] gives a maximal submodule $N$ of the left $A$-module $J^n$, so $J^n\mathrm{/}N$ is simple by [L3]. Since $J$ lies in every maximal left ideal of $A$, it annihilates every simple quotient of a finitely generated left module; in particular $J(J^n\mathrm{/}N)=0$. Hence $J^{n+1}=J{J}^n\subseteq N$, 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 still has finite length by [L1]. We prove by induction on its composition length that a finite-length module whose maximal submodules intersect trivially is semisimple. If $M=0$ there is nothing to prove. Otherwise choose a minimal nonzero submodule $S\le M$, 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 $M=S\oplus N$. The maximal submodules of $N$ correspond to the maximal submodules $S\oplus N^{\mathrm{\prime }}$ of $M$ that contain $S$, so their intersection in $N$ is again $0$; the induction hypothesis 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.