First, you need to know that the minimal polynomial of over F is for some (not a hard lemma)
Then if L is Galois over F, we must have in order to throw all conjugates of into K. But clearly is not real unless .
Sorry. I mixed up with something else. The ``lemma'' I stated before is not only unnecessary but also wrong. The correct version of the lemma needs the assumption that . The proof of the correct version needs Hilbert Theorem 90 using norms (or traces if n divides the characteristic of F), which is not hard but nontrivial to prove.
However, we don't even need this. is a root of x^n-a implies that all roots are of the form for some k (but not the converse, i.e. not all things of this form are the roots of x^n-a), since the minimal polynomial of must divide x^n-a. But use my idea before, that only two of the can be real. This implies that can only have at most two conjugates, hence the degree is <= 2