Let $T: X \to X$ an operator in the complex space X. Suppose that $T^k = I$ for some positive integer k. Show that $T$ is diagonalizable.
I know that $T^k - I = 0$ $\implies$ $p(T) = 0$ for $x^k -1 = p(x)$, so m(x) = $x^n -1$ is a minimal polynomial for n$\leq$k.
I'm a little bit stuck after that, i don't know if i'm going in the right direction.
The polynomial $p(x) = x^k - 1$ has $k$ distinct roots in $\Bbb{C}$ (i.e splits over $\Bbb{C}$ and each root has algebraic multiplicity $1$). Since the minimal polynomial $m_T(x)$ divides the annihilating polynomial $p(x)$, it follows that $m_T(x)$ also splits over $\Bbb{C}$, and each root has algebraic multiplicity $1$. This is equivalent to $T$ being diagonalizable (there are many proofs of this equivalence online and on this site as well).
By the way, it is not true that the minimal polynomial is of the form $x^n -1$ for some $n \leq k$. For instance, let $T=-\text{id}$, and let $p(x) = x^2 - 1$. Then, clearly $p(T) = 0$, but the minimal polynomial of $T$ is $m_T(x) = x+1$ (if what you wrote is true, the minimal polynomial would have to be either $x^2 -1$ or $x-1$, and this is false).