I'm asked the following question: Prove that $b$ is a primitive root modulo $p \implies$ the smallest positive exponent $e$ such that $b^e \equiv 1 \pmod p$ is $p - 1$.
I know that this could probably be shown easily with Fermat's Little Theorem, but it's posed to the reader before Fermat's Little Theorem is even discussed.
How should I approach this problem? Is there a simple argument for why this is true that doesn't require quite as much ingenuity as the proofs of Fermat's Little Theorem? (i.e., the ones that involve $(p-1)!$ or group theory)
In that case, look at all the powers of $b$, i.e. $b, b^2, b^3 \cdots$ modulo $p$ If $b^e\equiv 1 \mod p$, then the pattern repeats after $b^e$, i.e you get the same $e$ terms. So the totality of all numbers you can generate is $e+1$. But for a primitive root, you generate $p$ terms. So, $e=p-1$.
By the way the same holds for non-primes also. You just replace the condition by $e=\phi(p)$ using the same logic.