$T$ is diagonalizable if $T^n$ is identity for some $n$ Suppose $T$ is a linear operator on a $\mathbb{C}$-vector space $V$. Further, assume $T^n$ is the identity operator, for some $n$. Then, $T$ is diagonalizable.
I think there is a proof using Jordan theory, but I wish to find one without using it.

$$T^n=I\implies (T-I)(T^{n-1}+\ldots+T+I)=0\implies $$
the minimal polynomial of $\;T\;$ divides $\;T^n-I\;$, all of which roots are different (why?). Since you're working on an algebraic closed field this means all the roots of that minimal polynomial are in the field and thus the min. pol. factors in different linear factors $\;\iff T\;$ is diagonalizable.
