From a = 1 · a, it follows that

a + a + ··· + a	= (1 + 1 + ··· + 1) a = p · a = 0 · a = 0.
(p terms)

Newton's binomium leads to

(a+b)^p = _i=0^p ( ^p_i) aⁱ b^p-i = a^p+b^p

since for all i with 0 <i <p the binomial coefficient ( ^p_i) is divisible by p.

Since S has cardinality q, the set S\{0} contains exactly q - 1 elements, say a₁, a₂, ... , a_q-1.

Let a be one of these elements. Then multiplication by a is an injective map. Indeed, if ab = ac, then multiplying both sides with 1/a leads to b = c.

Hence, {a₁, a₂, ... , a_q-1} = S\{0} = {a a₁, a a₂, ... , a a_q-1}.

Consequently, the product P taken over all elements of S\{0} satisfies:

P = a^q-1P.

Indeed, a_i = (aa_i) = a^q-1 a_i.

The equality can be rewritten as

(1 - a^q-1)P = 0.

Since P is a product of nonzero elements, it is itself nonzero. Therefore, it is invertible, and we find 1 - a^q-1 = 0, and so a^q-1 = 1. Multiplication by a gives

a^q = a

as in the statement.

What remains is the proof for a = 0; but this case is trivial.