Let $R$ be a ring. Let $M$ be an indecomposable $R$-module with length $n$. Let $f\in\operatorname{End}(M)\setminus\operatorname{Aut}(M)$.
Show that $f^n=0$
What I have so far:
We can apply Fitting's Lemma, it tells us directly that $f$ is nilpotent.
If $f$ is nilpotent $\exists m$ s.t. $f^m=0$ and $\{f, f^2,...,f^{m-1}\}$ are all nonzero. Now we consider the chain $0 \subseteq ker f \subseteq....\subseteq kerf^{m-1}\subsetneq kerf^m =M$ where $kerf^{m-1}\subsetneq kerf^m$ holds because $f^{m-1}$ is assumed to be nonzero.
Suppose at some point we have $kerf^{i} = ker f^{i+1}$ for some $1 \leq i \leq m-1$. Then by some simple calculations we get that $ker f^{i}=kerf^{r}$ $\forall i \leq r \leq m$ and in particular $M=kerf^m=kerf^{i}$ which yields $f^{i}=0$. This contradicts our assumption that $\{f, f^2,...,f^{m-1}\}$ are all nonzero.
Hence we get a proper sequence $0 \subseteq ker f \subsetneq kerf^2 \subsetneq...\subsetneq kerf^m=M$. Since $M$ has length $n$ we can conclude $m \leq n$ and therefore $f^n=0$.
Is this correct?
Here is an idea, perhaps details are missing: Since $f$ is nilpotent, you have that $f^m=0$ for some $m$. Suppose $\{1,f,\dots,f^{m-1}\}$ are all nonzero. Then $Ker f^{i+1}\supsetneq Ker f^i$ and these are submodules. So you get an increasing sequence of proper submodules of $M$. Since the length of $M$ is $n$, it follows that $m\leq n$, and so $f^n=0$.