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I wish to prove the following:

If $p$ is an odd prime and $g$ is a primitive root modulo $p$, then either $g$ or $g+p$ is a primitive root modulo every power of $p$.

The only reference I can find to this is on page 26 of "A Course in Computational Algebraic Number Theory" by Henri Cohen (https://books.google.no/books?id=hXGr-9l1DXcC&pg=PP1&dq=A+Course+in+Computational+Algebraic+Number+Theory&ei=98TxSsRYjeSTBP7p6egL&redir_esc=y#v=snippet&q=primitive&f=false)

I'm struggling to understand the dense proof presented there. I understand all the steps and statements, with the exception of the last part of the following:

For any $m$ we have $m^p\equiv m\pmod{p}$, hence it follows that for every prime $l$ dividing $p-1$, $g^{p^{a-1}(p-1)/l}\equiv g^{(p-1)/l}\not\equiv 1\pmod{p}$. So for $g$ to be a primitive root, we need only that $g^{p^{a-2}(p-1)}\not\equiv 1\pmod{p^a}$.

Here is my understanding: $m^p\equiv m\pmod{p}$ is just Fermat's little theorem, and by induction it also follows that $m^{p^k}\equiv m\pmod{p}$. By inserting $a-1$ for $k$ and $g^{(p-1)/l}$ for $m$, we obtain $g^{p^{a-1}(p-1)/l}\equiv g^{(p-1)/l}\pmod{p}$. Further, since $g$ is a primitive root modulo $p$, $g^k\not\equiv 1\pmod{p}$ holds for all $k<p-1$, in particular for $k=(p-1)/l$. Going from there to say that it is sufficient that $g^{p^{a-2}(p-1)}\not\equiv 1\pmod{p^a}$ for $g$ to be a primitive root modulo $p^a$, is beyond me. Could anybody please help me understand this step?

The rest of the proof after that point is pretty straightforward, although it omits a lot of details.

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The order of $g$ mod $p^a$ has to divide $p^{a-1}(p-1)$ by Fermat-Euler, and you have shown you can't take out factors from $p-1$. So the order must be $p^j(p-1)$ for some $0\leq j\leq a-1$. Hence it suffices to impose the constraint $g^{p^{a-2}(p-1)}\not\equiv 1\pmod{p^a}$.

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