This problem is not very hard to prove. Especially if you read then end of page 387 and the beginning of page 388 (The proof is basically there).
First off: Suppose F is a subfield of GF(p^n) then F is isomorphic to GF(p^m) for some arbitrary m. By theorem 21.5 it follows that since m divides n then
n = [GF(p^n) : GF(p)]
n = [GF(p^n) : GF(p^m)]*[GF(p^m) : GF(p)]
n = [GF(p^n) : GF(p^m)]*m
And thus [GF(p^n) : GF(p^m)]= n/m
--Bakey 08:50, 30 November 2012 (UTC)
