$\bf Problem (a) $
Suppose that V is finite dimensional, and for each $v \in V$ there is some $k \geq 0$ (with k possibly depending on $v$) such that $T^k (v) = 0$. Prove that T is nilpotent.
Let $ \beta$ be a basis for V ie $ \beta = \{ v_1, \cdots , v_n \} $ then by assumption we have that $\exists k_i $ s.t $ T^{k_i}(v_i)=0 $ for each $v_i \in \beta $ then we let $ c = \max \{k_1, \cdots , k_n \} $ as any map from a basis exstends to a unique linear map we then have that $ T^{c}(v)=0 $ $\forall v \in V$ it then follows that T is nilpotent.
$ {\bf Problem (b)}$
Find an example to show that the statement in part (a) is not necessarily true if V is not finite dimensional.
if we consider $\mathcal P (x) $ then clearly if we take the differentiation map $D $ and define a basis $\beta = \{ v_1 , \cdots \}$ we know there exists a k s.t $ D^{k_i}(v_i) =0 $ the problem is that $ c = \max \{k_1, \cdots \} $ clearly is infinity how do i write this up formally?
You've done all the work already. In $P(x)$, the operator $D$ is not nilpotent, but for all $n$ $D^{n+1}(x^n)=0$, while $D^n(x^n)\ne0$.