Suppose , where for each . Prove that .
We can see, by an inductive argument, that has degree over for some . (In the chain of fields , each entry has degree 1 or 2 over its predecessor, and Theorem 14 in D&F applies.)
Suppose now that . Then , and we have the following diagram of field extensions.
Again using Theorem 14, we have that 3 divides , a contradiction. So in fact .