Algebraic closure of the field of two elements

I think I learned about this from a comment on MathOverflow.

Recall that the field of two elements \mathbf F_2 is the ring \mathbf Z/2\mathbf Z of integers modulo 2. In other words, it consists of the elements 0 and 1 with addition 1+1 = 0 and the obvious multiplication. Clearly every nonzero element is invertible, so \mathbf F_2 is a field.

Lemma. The field \mathbf F_2 is algebraically closed.

Proof. We need to show that every non-constant polynomial f \in \mathbf F_2[x] has a root. Suppose f does not have a root, so that f(0) \neq 0 and f(1) \neq 0. Then f(0) = f(1) = 1, so f is the constant polynomial 1. This contradicts the assumption that f is non-constant. \qedsymbol