prove that $ImT^k$ $\cap$ $KerT^k$ $= 0$ Let $T:V\rightarrow V$, be linear map such that $\dim V = n$ 
I am trying to prove that to all $k \ge n$:
$\mbox{im}T^k\cap\ker T^k$ $= \{0\}$.
I got stuck somewhere along the way,
I'm a linear algebra 1 student so if you could keep things simple I would appreciate it
Thanks!

Let $k \geq n$. There are two cases: Either $T^k \equiv 0$ or $rank T^k \ > 0$. First case is trivial as $ im T^k = \{0\}$. In the second case, we must have $rank T^l = rank T^{l+1}$ for some $l<n$, by the pigeonhole principle and the fact that rank of $T$ must weakly decrease by each iteration. This means $T$ is an isomorphism when restricted to $im T^l$. This should give you the result.
