There is a problem that has baffled my mind for a few days:
Let E be a n-dimensional vector space ($n\geqslant 2$); Linear transformation $T \in L(E)$ and basis ${e_1,e_2,...,e_n}$ for E exist such that $T(e_1)=e_2$, $T(e_2)=e_3$ , ... ,$T(e_{n-1})=e_n$ and $T(e_n)=0$. Is there any $S \in L(E)$ for $k\in \mathbb{N}, k\geqslant2$ suchthat $S^k=T$?
No. Or rather yes in the trivial case $n=1$, no if $n\ge2$. $S$ is nilpotent, so $S^n=0$. Now if $k\ge2$ there exists $m<n$ with $km\ge n$; hence $$T^m=S^{km}=0.$$But $T^m\ne0$ for $m<n$.
(Hmm, is it really true that if $k\ge2$ there exists $m<n$ with $km\ge n$? Sounds right... Ok, choose $m$ so $km\ge n$ and $k(m-1)<n$. Then $m<n$, because if $m\ge n$ then $k(m-1)\ge kn-k\ge 2n-2\ge n$ for $n\ge2$.)
Edit: Of course if $n=1$ then the conditions $T(e_1)=e_2$ and $T(e_n)=0$ are inconsistent. But in that case the first makes no sense, since there's no such thing as $e_2$; so I'm assuming that if $n=1$ the definition is $T=0$.