Let $A$ be an injective endomorphic operator on some functions space (wlog say $C^{\infty}(\mathbb{R})$). Then it cannot be nilpotent.
Does the above statement hold? I've tried to prove it like that:
Proof by induction. For $n=1$, this is just injectivity. Assume it holds for $n-1$, ie. $A^{n-1}$ in injective. Then $A^n f = 0$ is equivalent to $A(A^{n-1}f))=0$. As $A$ is injective, then $A^{n-1}f=0$, but as $A^{n-1}$ is injective aswell, we ge that $f=0$.
As every power of the operator $A$, ie. every $A^n$ is injective, A cannot be nilpotent.