If $a$ is a primitive root modulo $p$, then $(p-1) | ord(a)$ in $\mathbb{Z}/p^e\mathbb{Z}$ I'm a bit confused on a fact that my book has been using rather liberally without proof and I'm sure I'm missing something incredibly simple. Plainly stated, my question is if $a$ is a primitive root modulo $p$ so that
$$a^{p-1}\equiv 1 \pmod p$$
Then why is it that if the order of $a$ modulo $p^e, e > 1$ is $m$, then $p-1|m$? I don't understand this since I don't see any connection between the order of some element in $p$ versus it's order in $p^2, p^3, ....., p^e$ (this is also the "heart" of my question, what is the connection of orders of elements modulo $p, p^2, p^3, ...., p^e$?).
Furthermore, $\phi(p^e) = p^{e-1}(p-1)$ but I don't see why this necessarily shows that $$(p-1)\ |\ \text{ord}_{\mathbb{Z}/p^e\mathbb{Z}}(a)$$
Could someone please enlighten me on this?
 A: Max basically answered this question so just to reiterate what he posted in the comments, if $a$ is a primitive root modulo $p$ then let $m$ denote the order of $a$ in $\mathbb{Z}/p^e\mathbb{Z}$, then we have
$$a^m \equiv 1 \pmod {p^e} \implies p^e | a^m - 1 \implies p | a^m - 1 \implies a^m \equiv 1 \pmod p$$
But then $a^{p-1} \equiv 1 \pmod p$ since it is a primitive root modulo $p$, then $ord(a)_{\mathbb{Z}/p\mathbb{Z}} | m$ so indeed $p-1 | m$
A: This actually follows from a more general result.
Proposition: Let $G$ and $H$ be groups and $\varphi: G \to H$ be a homomorphism.  Then $\DeclareMathOperator{\ord}{ord} \ord(\varphi(g)) \mid \ord(g)$ for all $g \in G$.
Proof: Letting $m = \ord(g)$, then $\varphi(g)^m = \varphi(g^m) = \varphi(1) = 1$.  Then $\ord(\varphi(g)) \mid m$.
Now apply the proposition to your primitive root mod $p$ and the quotient map
$$
\pi: \mathbb{Z}/p^e \mathbb{Z} \to \frac{\mathbb{Z}/p^e \mathbb{Z}}{p\mathbb{Z}/p^e \mathbb{Z}} \cong \mathbb{Z}/p\mathbb{Z} \, .
$$
