Suppose $T_1,\ldots,T_{n+1}$ are pairwise commuting linear operators on an $n$-dimensional vector space $V$. Suppose $$T_1T_2\ldots T_{n+1}=0$$ Prove that in the above equation at least one of the factors can be removed without changing its validity.
What I can say so far is only that $T_1\ldots T_n=0$ implies that there exists $i\in \{1,\ldots,n+1\}$ such that $\mbox{Im}(T_{j+1})\subset\mbox{Ker}(T_j)$. However, I can't see how I'm supposed to use the fact that linear operators commute -- if they were diagonalizable, then I could say that they are simultaneously diagonalizable, but diagonalizability is nowhere mentioned.
I ran out of ideas. Any hints how to approach this problem?
$\DeclareMathOperator\Im{Im}$Consider the flag of subspaces $$0=U_0⊆U_2⊆⋯⊆U_{n+1}=V,$$ where $U_i=\Im(T_{i+1}T_{i+2}⋯T_{n+1})$ for $i=0,…,n$.
This is a flag since $T_{i+1}$ commutes with $T_{i+2}\cdots T_{n+1}$ so that \begin{align} U_i &= \Im(T_{i+1}T_{i+2}⋯T_{n+1}) \\&= \Im(T_{i+2}⋯T_{n+1}T_{i+1}) \subseteq \Im(T_{i+2}⋯T_{n+1}) = U_{i+1}. \end{align}
Note that we have $$U_i=T_{i+1}(U_{i+1}) \qquad\text{for $i=0,…,n$.}$$
Let us denote by $r_i=\dim(U_i)=\operatorname{rank}(T_{i+1}T_{i+2}⋯T_{n+1})$ the dimensions of the subspaces, then we get a weakly increasing sequence of integers $$0=r_0≤r_1≤⋯≤r_n≤r_{n+1}=n.$$ Hence, there must be an index $k≤n$ such that $r_k=r_{k+1}$. But this means that $U_k=U_{k+1}$ so that \begin{align} 0&=U_0=T_1T_2⋯T_k(U_k) \\&=T_1T_2⋯T_k(U_{k+1}) \\&=T_1T_2⋯T_kT_{k+2}⋯T_{n+1}(V). \end{align}
Hence, we may conclude that $$T_1T_2⋯T_kT_{k+2}⋯T_{n+1}=0.$$