Let $V$ be a finite dimensional vector space over $\mathbb{F}$ and let $T$ be a linear operator on $V$.Then $T$ is triangulable iff the minimal polynomial for $T$ is a product of the linear polynomial over $\mathbb{F}$.
The above theorem gives a proof to the cayley hamiltonian theorem over $\mathbb{C}$.[Hofmman and Kunze - Sec $6.4$]
I know that over $\mathbb{C}$ the characteristic polynomial is of the form $(x-c_1)^{r_1}\cdots (x-c_k)^{r_k}$ then the minimal polynomial over $\mathbb{C}$ is of the form $(x-c_1)^{e_1}\cdots(x-c_k)^{e_k}$[as minimal polynomial and characteristic polynomial have the same roots ] .
Also we know that trace ($[T]$) = $c_1.r_1 + \cdots c_k.r_k$.
Then $T$ is triangulizable and $[T]_{\beta}$ is a triangular matrix for some basis $\beta$ of $\mathbb{C}$.[from the above theorem].
Now $[T]_{\beta}$ is a triangular matrix with it's diagonal entries equal to the roots of the characteristic polynomial and also trace($[T]_{\beta}$) = $c_1.r_1 + \cdots c_k.r_k$
Hence $[T]_{\beta}$ satisfies its characteristic polynomial.Also since the minimal polynomial by definition is the smallest polynomial which is satisfied by $T$ then minimal polynomial divides the characteristic polynomial.
Is my reasoning okay?