Prove that $V = \ker(\phi^n) \oplus \text{image}(\phi^n)$
We consider the chains $$V\subset\phi(V)\subset\cdots\subset\phi^n(V)\subset\cdots$$ and $$\dim(V)\geq\dim(\phi(V))\geq\cdots\geq\dim(\phi^n(V))\geq\cdots$$
If $\dim(\phi^n(V))=1$, it is easy to prove.
If $\dim(\phi^n(V))\geq2$, there exist $k\leq n$ such that $\phi^k(V)=\phi^{k+1}(V)=\cdots$. Actually, $\phi:\phi^n(V)\rightarrow\phi^n(V)$ is an isomorphism.
- For any $a\in V$, there exists $b$ such that $\phi^n(a)=\phi^{2n}(b)$. Then we can prove $a=c+\phi^n(b)$ and $\phi^n(c)=0$.
- We assume $a\in\ker(\phi^n)\cap\text{image}(\phi^n)$. That is to say, $\phi^n(a)=0$ and $\phi^n(b)=a$. If $a\ne0$, we get $\phi^{2n}(b)=0$ and $\phi^n(b)\ne0$ which will lead $\dim(\phi^n(V))>\dim(\phi^{2n}(V))$. That is a contraction.
Look at an another question Show that $V = \mbox{ker}(f) \oplus \mbox{im}(f)$ for a linear map with $f \circ f = f$.
In your question, $\phi:\phi^n(V)\rightarrow\phi^n(V)$ is an isomorphism. Also $\phi^n:\phi^n(V)\rightarrow\phi^n(V)$ is an isomorphism.
In another question, $f=id: f(V)\rightarrow f(V)$ is an isomorphism.
Lemma Let $f:V\rightarrow V$ be a linear map with $f:f(V)\rightarrow f(V)$ being an isomorphism. Then we have $$V=\ker(f)\oplus\text{image}(f)$$
My answer is just a cleaned up version of the answer by gaoxinge.
$\newcommand{\inters}{\cap} $$\newcommand{\im}{\mathrm{im}} $The following chain of subspaces of the $n$-dimensional space $V\!$, $$ V\supseteq\phi V\supseteq\phi^2V\supseteq\cdots~, $$ cannot strictly decrease for ever, thus there is the least $k$ such that $\phi^kV=\phi^{k+1}V\!$, and then $\phi^kV=\phi^lV$ for all $l\geq k$. Set $U:=\phi^kV\!$. In the strictly descending chain $$ V\supset\phi V\supset\cdots\supset\phi^k V $$ the dimensions of subspaces decrease by at least one at each step, so we must have $k\leq n$. It follows that $\phi^{n+1}V=\phi^nV=\phi^kV=U$, so that $\phi\, U=U$, which means that the restriction of $\phi$ to $\phi_U\colon U\to U$ is an isomorphism. As you observed it suffices to prove that $\im(\phi^n)\inters\ker(\phi^n)=0$, so let us consider any $y$ in this intersection. Since $y\in\im(\phi^n)=\phi^nV=U$, and $\phi_U^ny=\phi^ny=0$, and $\phi_U^n\colon U\to U$ is an isomorphism, it follows that $y=0$, and we are done.