Is a non-zero Linear Operator, who's to the power of V's dimension is zero, diagonalizable?

Question

Given a linear Operator $N:V \to V$,that satisfies the following:

• $\exists v \in V$ (non-zero) s.t. for every $n\lt k$, $T^n(v)\neq0$.
• $\forall u\in V$, $T^k(u)=0$

is N diagonalizable?

I was given this question and I tried to approach it using the fact that if a Linear Transformation T has eigenvalues $\lambda$ then for $T^n$ an eigenvalue is $\lambda^n$, but it does not seem to work.

Thank you!