Discriminant of odd cyclic Galois extension is not a power of 2

Question

Is there a way to prove that, given a Galois cyclic extension $\mathbb{Q} \subset F$ with odd order, there exists a prime $p \neq 2$ such that $p|\Delta_{F}$ ? I'm actually trying to prove that the only Galois finite extensions in which $2$ is the only prime that ramifies have Galois group of even order. Proceeding by contraddiction I used Feit-Thompson to reduce to the case where $F$ is abelian over $\mathbb{Q}$ and then deduced that its Galois group is cyclic of odd order, but then I got stuck in relating this infomation with the condition that the discriminant must be a power of 2 (in order for it being the only ramified prime).