Please, check my proof that smallest natural number "e" where a^e=1(mod p) must divide (p-1)

[Boi]

First things first: I have posted my proof here because I don't know where to post questions about number theory. What I needed to proof definitely wasn't an advanced math theorem; just a simple task from an old book of mine.

Prove that the smallest number natural number $e$, satisfying the equation $a^e \equiv 1\pmod p$, must divide (p-1).
My proof:
Suppose the opposite: we assume that $e$ is the smallest natural number satisfying $a^e \equiv 1\pmod p$ (later I will not write mod p) but it does not divide $(p-1)$. Then, we can say $p-1=ke+r$, where $0 < r < e$. By Fermat's Little Theorem, it must be true that $a^{ke+r} \equiv 1$ or 1) $(a^e)^k*a^r \equiv 1$. By our assumption, $a^e \equiv 1$, which also means that $(a^e)^k \equiv 1$. Knowing this, we can rewrite 1) as $a^r \equiv 1$. We definitely know $r \not = 0$ and $r < e$. We've just arrived at contradiction: $e$ was assumed to be the smallest natural number satisfying our equation but we just have found a smaller one. So, $e$ must divide $(p-1)$.

 

[blamocur]

Looks good, but a) do you need to mention that $p$ is prime? and b) that $a>1$ ?

 

[Boi]

Looks good, but a) do you need to mention that $p$ is prime? and b) that $a>1$ ?

Oh, yeah, I totally forgot that. Yes, $p$ is prime and $a > 1$. $p$ also must not divide $a$. This exercise is given in the textbook after chapter explaining Fermat's Little Theorem.