Showing that $V=\operatorname{ker} f^{p} \oplus \operatorname{Im} f^{p}$ where $\operatorname{Im} f^{p} =\operatorname{Im} f^{p+1}$ This is a problem from Blyth and Robertson's Further Linear Algebra:

If $V$ is a finite dimensional vector space and $f: V\to V$ is a
linear mapping, prove that there is a positive integer $p$ such that
$\operatorname{Im} f^{p}=\operatorname{Im} f^{p+1}$ and deduce that
$$(\forall k\ge 1)\; \operatorname{Im} f^{p}=\operatorname{Im}
 f^{p+k}, \ker f^{p}=\ker f^{p+k}$$ Show also that $V=\operatorname{Im}
 f^{p} \oplus \ker f^{p}$.

My attempt:
Let $\displaystyle m=\min_{k\in\mathbb{N}} (\dim f^{k}(V))$. Then there must be a $p\in \mathbb{N}$ such that $m=\dim f^{p}(V)$. It is easy to see that $f^{p+k}(V)\subset f^{p}(V)$, so,  $m=\dim f^{p}(V)\ge\dim f^{p+k}(V)$ and by definition of $m$, $\dim f^{p+k}(V)\ge m$ for any $k\in \mathbb{N}$. So, we have that $\dim f^{p+k}(V)=m$ for any $n\in \mathbb{N}$. Hence, we have $f^{p+k}(V)=f^p(V)$.
Also, we have that $\ker f^{p}\subset \ker f^{k+p}$. By rank nullity theorem we have $\dim \ker f^p =\dim \ker f^{k+p}$. Hence $\ker f^{p}=\ker f^{p+k}$.
Let $v\in V$. Then $f^{p}(v)\in f^{2p}(V)$ because $f^{p}(V)=f^{2p}(V)$. Thus, $f^{p}(v)=f^{2p}(w)$ for some $w\in V$. Thus, $f^{p}(v-f^{p}(w))=0$. Hence $v-f^{p}(w)\in \ker f^{p}$. Hence, we can write $v=v_0+f^{p}(w)$ where $v_0\in \ker f^{p}$. This shows that $V=\operatorname{Im} f^{p} + \ker f^{p}$. Now, by the dimension formula and rank nullity theorem, $\dim (\ker f^{p} \cap \operatorname{Im} f^{p})=\dim \ker f^{p} + \dim \operatorname{Im} f^{p} - \dim V=0$. So, $\ker f^{p} \cap \operatorname{Im} f^{p}=\{ 0 \}$. Hence  $V=\operatorname{Im} f^{p} \oplus \ker f^{p}$.

Is this proof correct? Is there a more direct way to do this?

Your proof seems correct to me. If $N_i$ denotes the kernel of $f^i$ then the chain $\{0\} = N_0 \subseteq N_1 \subseteq \cdots$, must eventually stabilise, because there are at most $d$ strict inclusions, where $d = \dim V$. Likewise, if $R_i$ is the range of $f^i$, then $V = R_0 \supseteq R_1 \supseteq \cdots$ must stabilise for the same reason.
For what it's worth, the decomposition of $V$ that you find is sometimes called the Fitting decomposition of $V$ with respect to $f$. That is, $V$ decomposes as a direct sum $N \oplus R$ such that $f|_{N}$ is nilpotent and $f|_{R}$ is invertible. Really cool!
