Finite characteristically simple groups are direct products of isomorphic simple groups

By [A1], choose a minimal normal subgroup $N⊴G$.

Let $N_1,\dots ,N_r$ be an irredundant family of automorphic images of $N$ that generates the subgroup $H$ generated by all automorphic images. By [A2], the subgroup $H$ is characteristic in $G$, so $H=G$ because $G$ is characteristically simple and $N\mathrm{\ne }1$.

Fix $i$ and put $P_i=⟨N_j:j\mathrm{\ne }i⟩$. The intersection $N_i\cap P_i$ is normal in $G$: both factors are normal, and the intersection is preserved by conjugation. Minimality of $N_i$ makes this intersection either $1$ or $N_i$. The latter would put $N_i$ inside the subgroup generated by the other images, contradicting irredundancy. Hence $N_i\cap P_i=1$ for every $i$. Together with step 2.1, [A2], and [L1], this makes $G=N_1\times \cdots \times N_r$ an internal direct product.

Let $K⊴N_i$. Since the other direct factors centralize $N_i$, conjugation by them fixes $K$, while conjugation by $N_i$ preserves $K$ by normality. Step 3.1 says these factors generate $G$, so $K⊴G$. Minimality of $N_i$ gives $K=1$ or $K=N_i$; thus $N_i$ is simple. All $N_i$ are automorphic images of $N$, so they are pairwise isomorphic to one finite simple group $T$. Therefore $G\cong T^r$.