Endomorphism Rings of finite length Modules are semiprimary
Added later:
Forget the reduction to linear algebra! There is a short and elementary proof which directly shows the original statement. It appears as an exercise in Bourbaki's “Éléments de mathématique” (Algèbre, chapitres 8, ex. 2). Here is my vague translation:
Claim: Let $M$ be a right $R$-module of finite length $n$, and let $\mathcal{F} \subset \mathrm{End}_R(M)$ be a nonempty set of nilpotent endomorphisms which is closed under multiplication (e.g. the Jacobson radical in the above setting). Then $\mathcal{F}^n = 0$, meaning that the composition of any $n$ endomorphisms of $\mathcal{F}$ is zero.
Proof: As in my proof given below, we argue by induction on the length of $M$ to construct a composition series of $M$ $$ 0 = M_0 < \dots < M_n = M, $$ such that $\mathcal{F} M_{i+1} \subseteq M_i$ for all $i$. If $\mathcal{F} = 0$ there is nothing to show. Otherwise, we have $l(M) \geq 2$, and it suffices to prove the existence of a proper $\mathcal{F}$-invariant submodule $0 \lneq N \lneq M$.
Let $0 \neq f \in \mathcal{F}$ such that $fM$ has minimum length among all nonzero elements of $\mathcal{F}$. It can be shown in the same way as below (see the lemma) that $fgf = 0$ for all $g \in \mathcal{F}$. If $\mathcal{F}f = 0$ then we are done by setting $N=fM$. If $\mathcal{F}f \neq 0$ we set $$N = \sum_{g \in \mathcal{F}} gfM.$$ This is a nontrivial $\mathcal{F}$-invariant submodule of $M$. By construction, we have $fN = \sum_{g \in \mathcal{F}} fgfM = 0$, so $N$ must be proper. $\square$
Original Answer:
Finally, I think I've found a proof. To begin with, note that the original problem appears in Lam's ”first course in noncommutative rings” as an exercise (3.24). Lam suggests to reduce the nilpotency statement to the problem, whether any (multiplicative) nil subsemigroup of a semisimple ring is nilpotent. I already explained how the reduction works in Edit 2 and Edit 3. Afterwards, using the Artin-Wedderburn structure theory, it is easy to reduce further to a linear algebra problem over division rings. In the end it remains to show the following:
Claim: Let $V$ be an $n$-dimensional vector space over a division ring $D$, and let $\mathcal{F} \subset \mathrm{End}_D(V)$ be a nonempty set of nilpotent endomorphisms which is closed under multiplication. Then $\mathcal{F}^n = 0$, meaning that the composition of any $n$ endomorphisms of $\mathcal{F}$ is zero.
Proof: It suffices to show that $\mathcal{F}$ is triangularizable, i.e. there are subspaces $$ 0 = V_0 \leq V_1 \leq \dots \leq V_n = V, $$ such that $\mathcal{F}V_{i+1} \subseteq V_i$ for all $i$. If $\mathcal{F}$ consists only of the zero endomorphism, there is nothing to show. If $\mathcal{F}$ is nontrivial (which already implies $n \geq 2$) we prove the existence of a proper $\mathcal{F}$-invariant subspace $0 \lneq W \lneq V$. Everything else follows by induction by considering $W$ and $V/W$.
Let $r \in \mathbb{Z}$ be the minimum rank of a nonzero element of $\mathcal{F}$. We define $$ \mathcal{F}_0 = \{ f \in \mathcal{F} : \mathrm{rank}(f) \leq r \}. $$ $\mathcal{F}_0$ is a left ideal of $\mathcal{F}$ in the sense that $\mathcal{F} \cdot \mathcal{F}_0 \subseteq \mathcal{F}_0$. For that reason, the (nontrivial) abelian group $W$ spanned by $\mathcal{F}_0 V$ is an $\mathcal{F}$-invariant subspace of $V$. We claim that $W$ is a proper subspace of $V$. But first we need a lemma:
Lemma: Let $f \in \mathcal{F}_0$, $g \in \mathcal{F}$. Then $f^2 = fgf = 0$.
Proof: Of course there is nothing to show if $f=0$, so let $f \neq 0$. Since $f$ is nilpotent, we have $\mathrm{rank}(f^2) < \mathrm{rank}(f) = r$, so $f^2 = 0$. Suppose $fgf \neq 0$. Then $r \leq \mathrm{rank}(fgf) \leq \mathrm{rank}(f) = r$. A dimension argument shows that $fg$ restricts to an automorphism of $\mathrm{Im}(f)$ which is impossible, since $fg$ is nilpotent. $\square$
Proof of $W < V$: Suppose $W = V$. Let $\{f_1, \dots, f_m\} \subseteq \mathcal{F}_0$ be a minimal set such that $$V = \mathrm{Im}(f_1) + \dots + \mathrm{Im}(f_m).$$ Because of the lemma, we may assume without loss of generality that $f_1f_i = 0$ for all $i$: If $f_1f_i \neq 0$ for some $i$, we can simply replace $f_1$ by $f_1f_i$, and the image won't change. The lemma guarantees that this process terminates after finitely many steps, as it ensures that any product of elements of $\mathcal{F}_0$ with repeatedly occurring factors is zero. Now by multiplying both sides of the equation by $f_1$ from left, we arrive at the follwing contradiction: $$ \mathrm{Im}(f_1) = f_1 V = f_1\mathrm{Im}(f_1) + \dots + f_1\mathrm{Im}(f_m) = \mathrm{Im}(f_1f_1) + \dots + \mathrm{Im}(f_1f_m) = 0. $$ $\square$
I hope there is no mistake this time. Comments are highly appreciated!