Odd-order p-group implies every irreducible representation has Schur index one

Statement
Suppose $$p$$ is an odd prime and $$G$$ is a finite $$p$$-group, i.e., a group of prime power order where the underlying prime is $$p$$. Then, every irreducible representation of $$G$$ over a splitting field has fact about::Schur index 1, i.e., every irreducible representation can be realized over the field generated by the character values of the representation.

Related facts

 * The statement is not true for $$p = 2$$. An example is the faithful irreducible representation of quaternion group, which is a degree two representation whose character takes values in $$\mathbb{Q}$$ but where the representation cannot be realized over $$\mathbb{Q}$$ but is realized over a suitable imaginary quadratic extension of $$\mathbb{Q}$$, such as any extension of the form $$\mathbb{Q}(\sqrt{-m^2 - 1})$$.
 * The statement is not true for odd-order groups in general. In particular, there are counterexample odd-order groups (which are, by the odd-order theorem, finite solvable groups) that are not nilpotent and where one or more of the irreducible representations has Schur index greater than one. The smallest example is nontrivial semidirect product of Z7 and Z9.