Let $V$ be the space of all $f(t) \in K[t]$ with $\mathrm{deg} f \leq n-1$ and let $\psi: V \to V$ with $\psi(f) = f'$. Further $\mathrm{char}(K) = 0$.
Then $\psi$ is nilpotent. Since one can take the derivative over and over again until $f^{(m)} = 0$. Further more $V$ is cyclic since $\frac{t^{n-1}}{(n-1)!}, \left(\frac{t^{n-1}}{(n-1)!}\right)', ..., t, 1$ form a basis of $V$.
But why is it important that $\mathrm{char}(K) = 0$?
If $\operatorname{char}(K) \neq 0$ the term $\frac{t^{n-1}}{(n-1)!}$ is nonsense but how to prove that there can't be another cyclic base if $\operatorname{char}(K) \neq 0$?
Would it work if $\operatorname{char}(K) \geq (n-1)!$?