Is my solution correct, and is there a better one? This is from Dummit and Foote, section $11.1$. I would please like to know if my proof is right, if it can be improved, and if there is some better proofs.

$6)$ Let $V$ be a vector space of finite dimension. If $\phi$ is any linear transformation from $V$ to $V$, prove there is an integer $m$ such that the intersection of the image of $\phi^m$ and the kernel of $\phi^m$ is $\{0\}$. 

Proof: We have that $V \supseteq \phi(V) \supseteq \phi^2 (V) \supseteq \phi^3(V) \cdots$
Therefore we also have $\dim V \ge \dim \phi(V) \ge \dim \phi^2(V) \ge \cdots$.
Since the dimension is finite, we must have, after some $m$, that $\phi^m(V) = \phi^{m+1}(V) = \phi^{m+2}(V) = \cdots$.
Therefore, $\phi$ restricted to $\phi^m(V)$ is onto $\phi^m(V)$, and this implies that it is also injective. Therefore it's kernel is $\{0\}$. But this is the same as $\ker \phi^{m+1} \cap \text{range } \phi^{m+1}$, and the proof is complete.
 A: As I said in a comment I don't think this is quite enough. You get to
"We have after some $m$
$$
\phi^m(V) = \phi^{m+1}(V) = \phi^{m+2}(V) = \cdots
$$
Therefore, $\phi$ restricted to $\phi^m(V)$ is onto $\phi^m(V)$, and this implies that it is also injective. Therefore its kernel is $\{0\}$".
So you  have proved that $\ker\phi |_{\phi^m(V)}=\{0\}$; that is $\ker\phi\cap\phi^m(V)=\{0\}$. This isn't quite what you are asked to prove. But you can adapt it slightly as follows.
We have after some $m$
$$
\phi^m(V) = \phi^{m+1}(V) = \phi^{m+2}(V) = \cdots=\phi^{2m}(V)
$$
Therefore, $\phi^m$ restricted to $\phi^m(V)$ is onto $\phi^{2m}(V)=\phi^m(V)$, and this implies that it is also injective. Therefore its kernel is $\{0\}$.
So we have proved that $\ker\phi^m |_{\phi^m(V)}=\{0\}$; that is $\ker\phi^m\cap\phi^m(V)=\{0\}$.
[I have one other suggestion. It helps (especially when it comes to revision) to make clear which big results you are using on the way; this helps place the result in the Big Picture. I think in this question you have used (silently) at least three. (I) Every non-empty set of natural numbers has a minimum element. (II) Every LI set of vectors can be expanded to a basis. (This is what lets you use at various places " Suppose $X\leqslant Y$. Then $X=Y$ iff $\dim X=\dim Y$.") (III) The Rank-Nullity Theorem.]
