By Newton's binomium, and the fact that all but the two extreme binomial coefficients are zero,

(x + y)p =

m

p m

= xp + yp.

For the proof regarding the binomial coefficients, see Exercise 11 of Section 2.6.

To prove the equation with q instead of p, we can use induction on the number a such that q = p^a. Above we have estabished the case a = 1. Suppose we have dealt with the case a-1. Then, using the induction hypothesis and q/p = p^a-1, we find

(x + y)^q = ((x + y)^q/p)^p = (x^q/p + y^q/p)^p = x^q + y^q.

We have to show

1. (xy)^q = x^q y^q;
2. (x + y)^q = x^q + y^q;
3. 0^q = 0,       1^q = 1.

The first and third statement are obvious. The second has just been proved in Part 1.

This is immediate from Part 2 and the fact that x^p = x for x Z/pZ (known as Fermat's little theorem).

Since x -> x^q is a morphism by Part 2, the subfield result follows from the lemma on the previous page.

Finiteness follows from the previous lemma.

Put M = {x K | x^p = x}. By Part 4, M is a subfield of K. Clearly, the smallest subfield L of K is contained in M, so we only need show that M has no more than p elements. But elements of M are zeros of the polynomial X^p - X, and so there are at most p solutions by the previous lemma.