How can I prove that for a nilpotent endomorphism $\xi: V \to V$ with $\xi^k(V) = 0$ and $k$ being the smallest natural number to do so:
$$\{0\} = \xi^k(V) \subsetneq \xi^{k-1}(V) \subsetneq \dots \subsetneq \xi(V) \subsetneq V $$
and deduce from that fact that $k \leq n = dim(V)$? Intuitively, it is clear that less and less basis vectors are spanning the $\xi$-invariant spaces but I have no clue how to show that in a formally correct fashion. As soon as this is showed, the corrolary of $k \leq n$ is clear I think.
Hint
For every $m$ the inclusion $\xi^{m+1}(V)\subset \xi ^m(V)$ is trivial. So it suffices to get the desired result to prove that $$\xi^{m+1}(V)=\xi^m(V)\implies \xi^{p}(V)=\xi^m(V),\;\forall p\ge m$$ which is done by induction using that $$\xi^{p+1}(V)=\xi(\xi^p(V))=\xi(\xi^m(V))=\xi^{m+1}(V)=\xi^m(V)$$