$\exists n\in\mathbb{N}$ such that $T_1(\ldots(T_n(V)\ldots)=\{0\}$, for $T_i$ nilpotent operators and $dimV<\infty$.

Question

I'm studying Nilpotent Operators and based on this question Nilpotent Operators Linear Algebra and the variation of the same question proposed by this (accepted answer) here I've found difficult to prove or give a counterexample to the following:

Let $V$ be a finite-dimensional vector space. Is there a finite index $n>0$ such that every composition of $n$ nilpotent linear operators $T_i\colon V \to V$ results on the zero operator, that is, $T_1(\cdots (T_n(x))\cdots)=0,\forall x\in V$? The operators need not to commute.

The only requirements on the field is that its characteristic is not 2. Even algebraically closed fields are of interest. Any references would also be of great help.

Answer 1

Let $$ A=\pmatrix{0&0\cr1&0\cr},\qquad B=\pmatrix{0&1\cr0&0\cr} $$ Then $A^2=B^2=0$ but $$ AB=C=\pmatrix{0&0\cr0&1\cr} $$ is not, so $C^n$ is a nonzero product of $2n$ nilpotents.