$T^m(v)=v$ implies $T$ diagonalizable 
Suppose $V$ is a vector space over $\Bbb C$, and let $T:V\to V$ be a linear transformation. such that there exists $m\in \Bbb N$ such that $T^m(v)=(v)$. Prove that $T$ is diagonalizable.
Proof: Fix a basis $v_1,\dots,v_n$ of $V$. Let $m_1,\dots,m_n$ be such that $T^{m_i}v_i=v_i$. Let $n=\text{lcm} \{n_i:1\le i \le d\}$. Then $T^n(v)=v$ for every $v\in V$, so over $\Bbb C$, the characteristic polynomial has distinct roots.

I can't understand why it has distinct roots.

The book is using sloppy language.  Of course, the characteristic polynomial need not have distinct roots.  The author meant to say that the minimal polynomial has distinct roots, which you can see because $x^n-1$ has no repeated roots, the roots being the $n$ $n^{th}$ roots of unity.  (And since T satisfies $x^n-1=0$ its minimal polynomial must divide $x^n-1$.)
Note that the book is also sloppy in not quantifying $v$ and $m$ in the statement of the problem.  The author really should have said either "for every $v$ there is some $m$..." or "there is some $m$ such that for every $v$...".  Even though the problem works in either of those formulations, it prevents the misreading "For some $v\ne 0$ and some $m\ge 1$..." which would of course not work.
