$\Rightarrow$ $(-\zeta)^{2p}=(-\zeta^p)^{2}=1$ and if $i \in[{1,2p-1}]$ there is $k \in[0,p-1]$ such that $i=2k+1$ if is odd and $2k$ if is even so $(-\zeta)^{i}=(-\zeta)^{2k}=(\zeta^k)^{2}\neq 1$ since $\zeta$ is p root and $ p> 2$ ($p$ is prime and odd).
but i dont know if this is right, if it is then the odd case will follow to.
My question is indeed in how to prove $\Leftarrow$, if $(-\zeta)^{i}\neq 1$ $\forall i \in[{1,2p-1}]$, how can i use this to prove $(\zeta)^{i}\neq 1$ $\forall i \in[{1,p-1}]$? the negative sign is what confusing me
Answering your question on the reverse implicatuon: If z is a 2p primitive root then $z^p=-1$ since -1 is the only primitive square root. Then $(-z)^p=1$ since $p$ is odd. And $(-z)^j\ne 1$ for $0<j<p$ lest $z^{2j}=1$, contradicting that $2p$ is the least positive power of z that equals 1.
On the forward implication, if some power k of (-z) less than 2p were unity, then (a) if that power is p then since p is odd $z^p=-1$, contradiction; (b) otherwise k and therefore 2k is prime to p and $z^{2k}=1$, contradiction.
I used z for $\zeta$ throughout to save keystrokes.