$M$ is an $R$-module with composition series. Show that for any $R$-homomorphism $f: M\rightarrow M$ there is a natural n such that $M = \operatorname{Im}f^{n}\oplus\ker f^{n}$.
We know that $M$ has composition series, which means that $M$ is Artinian and Noetherian. Now the kernel of any $R$ - module homomorphism is always a submodule, so we can consider the following ascending chain of kernels
$\ker f \subset \ker f^{2} \subset \ker f^{3} \subset \cdots \subset \ker f^{n} \subset \cdots$
that must eventually become constant by the Noetherian condition so we may suppose that there is a natural $n$ such that
$\ker f^{n} = \ker f^{n+1} = \ker f^{n+2}= \cdots $
I also thought to show that $\ker f^{n} \cap \operatorname{Im} f^{n} = \left \{ 0 \right \}$ so $M = \operatorname{Im}f^{n}\oplus \ker f^{n}$ from the definition of a vector space being the direct sum of two subspaces.