for an endomorphism $f$ over an Artinian module

Question

For an endomorphism $f$ over an Artinian Module $M$ over a ring $R$ i have to show that $M = f^n(M) + ker(f^n)$, we get quickly from one of the isomorphism theorems that $f^n(m) \simeq M/ker(f^n)$ can i from this get my conclusion?

Answer 1

$im(f)\supseteq im(f^2) \supseteq ...\supseteq im(f^n)\supseteq ...$ and since $M$ is Artinian there is some $n$ s.t. $im(f^n)=im(f^{n+1})$

Let $x\in M$. Then $f^n(x)\in im(f^n)=im(f^{2n})$ hence $f^n(x)=f^{2n}(y)$ for some $y\in M$. Hence $x-f^n(y)\in ker(f^n)$ and $x=f^{n}(y)+(x-f^{n}(y))\in Im(f^n)+ker(f^n)$