Show that every finite field is of prime-power order, that is, it has a prime-power number of elements.
Detailed Proof:
Step 1: Every finite field has a prime characteristic and contains a prime field.
Let F be a finite field. Since F is finite, its characteristic cannot be 0, as a field of characteristic 0 would contain a copy of `QQ` , which is infinite. Let F have characteristic p.

Input Explanation:.
Prime Subfield: Every field F has a smallest subfield, called its prime subfield. This subfield is either isomorphic to `QQ` (if the characteristic is 0) or to `F_p` (for prime p , if the characteristic is p).
Consider the set of elements { 1 , 1 + 1 , 1 + 1 + 1 , ... } in F. Since F is finite, this sequence must eventually repeat. This means there exists distinct positive integers m > n such that
1 + ... + 1 (m times) = 1 + ... + 1 (n times). This implies 1 + ... + 1 (m - n times) = 0. The smallest positive integer k such that 1 + ... + 1 (k times) = 0 is the characteristic of F.
Let this characeristic be p.
We claim that p must be a prime number. Suppose p = ab for some integers 1 < a, b < p.
Then 1 + ... + 1 = 0 (ab times). This can be written as
(1 + ... + 1)(a times) `*` (1 + ... + 1)(b times) = 0. Since F is a field, it has no zero divisors. Thus either (1 + ... + 1)(a times) = 0 or (1 + ... + 1)(b times) = 0. This contradicts the minimality of p as the characteristic. Therefore , p must be a prime number.
P = {0 , 1 , 1 + 1 , ... , 1 + ... + 1} forms a subfield of F. This subfield is isomorphic to `ZZ_p` , the field integers modulo p. This subfield P is called the prime field of F.