Skeptical about an Elementry point concerning polynomials of linear operators

It does not hold.

Let $T = \pmatrix{0 & 1 \\ 0 & 0}$.

We have that $T^2 = 0$ so in particular $T^2 e_2 = 0$. However, $Te_2 = e_1 \ne 0$.

The answer is no if you insist on using the same $v$, as @mechanodroid has told us.

But it does imply that for at least one $j$ we have $(T-\lambda_jI)w = 0$ for some $w \in V$ with $w \ne 0$.

Indeed, if $(T-\lambda_2I)\cdots(T-\lambda_mI)v \ne 0$, we can take $j=1$ and $w=(T-\lambda_2I)\cdots(T-\lambda_mI)v$. Otherwise, we repeat the process with $(T-\lambda_2I)\cdots(T-\lambda_mI)v =0$.