Prove an upper bound for the multiplicative order of a congruence

Question

This is a problem from elementary Number Theory. It's the only one I couldn't figure out and it's bothering me.

Definition: Let a and n be natural numbers with (a, n) = 1. The smallest natural number k such that a^k ≡ 1 (mod n) is called the order of a modulo n and is denoted ord_n(a).

Prove: Let a and n be natural numbers with (a, n) = 1. Then ord_n(a) < n.