Let $V$ a finite dimensional vector space over $\mathbb{C}$ and $T:V \to V$ a linear operator. Show that $T^{n+1} = T \iff T^{2n} = T^n$ and T is diagonalizable.
My attempt:
$(\Rightarrow)$
If $T^n= T$, then $p(T) = 0$, where $p \in \mathbb{C}[X]$ is given by $p(x) = x^n - x$. Now take the polynomial $q(x) = x^{n-1}p(x) = x^{2n} - x^n$. Since $p$ divides $q$ we have $q(T) = 0$, that is, $T^{2n} = T$.
$(\Leftarrow)$
Suppose that $T^{n+1} \neq T$, then applying $T^{n-1}$, we have $T^{2n} \neq T^n$ (Is that correct?).
Since $p(T)=0$ and $p(x)$ we have $m_T| p(x)$. However, $p$ has distinct roots then $p = m_T$ and $T$ is diagonalizable and it eigevalues is $0,1,\omega,...,\omega^{n-1}$, where $\omega^i$ is the roots of $x^n-1$.
The premise of the forward implication is that $T^{n+1}=T$, not $T^n=T$. Also, you haven't shown that $T$ is diagonalisable.
In proving backward implication, you claim that $T^{n+1}\ne T$ implies that $T^{2n}\ne T^n$. This is not correct. E.g. when $n=2$ and $T=\pmatrix{0&1\\ 0&0}$, we have $T^{n+1}=T^3=0\ne T$ but $T^{2n}=T^4=0=T^2=T^n$. Note that $T$ is not diagonalisable in this counterexample.
Hints for proving the statement in question:
$(\Rightarrow)$ Note that $x^{n+1}-x$ can be factorised into $n+1$ distinct linear factors. Now the minimal polynomial of $T$ must divide any annihilating polynomial.
$(\Leftarrow)$ As $T$ is diagonalisable, you may assume that $Tv_i=\lambda_iv_i$ for some eigenbasis $\{v_1,v_2,\ldots,v_m\}$ of $V$.
$(\Leftarrow)$ Alternatively, since $T$ is diagonalisable, its minimal polynomial $m_T(x)$ is a product of distinct linear factors. However, as $x^{2n}-x^n$ annihilates $T$, $m_T(x)$ must divide $x^{2n}-x^n=x^n(x^n-1)$. Therefore $m_T(x)$ divides $x(x^n-1)$.