The context.
An endomorphism $u\in \mathcal L(E)$ is called nilpotent if, and only if,
$$\exists n\geqslant 0,\quad \forall x\in E,\quad u^n(x)=0.\qquad (\star)$$
For example,
$$\begin{matrix}D\colon& \mathbb R_{\leqslant n}[X] & \to &\mathbb R_{\leqslant n}[X] \\ & P&\mapsto &P'\end{matrix}$$
(where $\mathbb R_{\leqslant n}[X]$ stands for polynomials of $\mathbb R[x]$ of degree $\leqslant n$) is nilpotent since
$$\forall P\in \mathbb R_{\leqslant n}[X],\quad D^{n+1}(P)=0.$$
The question.
If we consider this time
$$\begin{matrix}D\colon & \mathbb R[X] & \to &\mathbb R[X] \\ & P&\mapsto &P'\end{matrix}$$
it is not nilpotent anymore since it only verifies
$$\forall P\in \mathbb R[X],\quad \exists n\geqslant 0,\quad D^n(P)=0.$$
Does this type of endomorphism (where the quantifiers have been exchanged in the definition $(\star)$) have a name?
This is called a locally nilpotent operator, here is a link from another math.stackexchange question (I couldn't find official source on this).