This question is a follow-up to my other question here: If $A$ is invertible and $A^n$ is diagonalizable, then $A$ is diagonalizable.
My problem is as follows:
Given a vector space $V$ over $\mathbb{C}$, and a linear operator $T$ on $V$, show that if $T$ is invertible and $T^k$ is diagonalizable for some $k\geq2$, then $T$ is diagonalizable.
My attempt at a solution:
Because the vector space is over $\mathbb{C}$, $T$ has an eigenvalue (by the fundamental theorem of algebra). Let the eigenvalues of $T$ be denoted by $\lambda_1,\lambda_2,\cdots,\lambda_n$, for some $n\geq1$. Then, observe that for some eigenvector $v$, $$Tv=\lambda_i v\Rightarrow\lambda_i (Tv)=\lambda_i(\lambda_i v)\Rightarrow T(\lambda_i v)=\lambda_i^2 v\Rightarrow T(Tv)=\lambda_i^2 v\Rightarrow T^2v=\lambda_i^2 v.$$ This argument can be inductively continued to show that the eigenvalues of $T^k$ are $\lambda_1^k,\lambda_2^k,\cdots,\lambda_n^k$. Because $T^k$ is diagonalizable, these eigenvalues are distinct (a proof that I'm omitting for brevity here). Because the $\lambda_i^k$'s are distinct, it follows that the $\lambda_i$'s are distinct. Therefore, $T$ is diagonalizable (again based on the proof that I'm omitting).
Here are my questions:
(a) First and foremost, is this a valid proof? I'm skeptical because it is so much simpler than the proof I've linked (which is in terms of matrices, not linear operators, but that shouldn't make a significant difference). I also haven't (explicitly) used the fact that $T$ is invertible, so I feel something is missing.
(b) It seems like this problem can be approached using Jordan canonical form, but I'm struggling to do so. Any suggestions would be appreciated.