As it's indicated in the title, I have to show that
$x$ generator of $(\mathbb{Z}/p^2\mathbb{Z})^* \implies$ $x$ generator of $(\mathbb{Z}/p^n\mathbb{Z})^*$, $p \neq 2$.
I'm not sure of my proof, it's why I post it here and I hope you will be able to enlighten me if it's okay or not ! :)
My proof :
We know that ($\mathbb{Z}/p^n\mathbb{Z})^*$ is cyclic. Let's $\overline{q}$ be a generator of $(\mathbb{Z}/p^n\mathbb{Z})^*$. So $\overline{x} = \overline{q}^m$, $m \in \mathbb{Z}$. But $\overline{q}$ is as well generating ($\mathbb{Z}/p^2\mathbb{Z})^*$, and we have : $x = q^m + kp^n$, $k \in \mathbb{Z}$. So, as $n \geq 2$, we have : $x = q'^m [p^2]$ ($\overline{x} = \overline{q}^m$, but this time mod $p^2$).
On the other side, we know that $\mathrm{ord}(\overline{x}) = p(p-1)$ over $(\mathbb{Z}/p^2\mathbb{Z})^*$ and $\overline{q}$ is generating $(\mathbb{Z}/p^2\mathbb{Z})^*$, so we should have $\gcd(m,p(p-1))=1$ (otherwise $x$ is not a generator over $(\mathbb{Z}/p^2\mathbb{Z})^*$).
From here, as $\gcd(p,p-1)=1$, we have $\gcd(m,p)=1, \gcd(m,p-1)=1$, and then $\gcd(m,p^n(p-1))=1$. So, $\mathrm{ord}(\overline{x}) = \mathrm{ord}(\overline{q}) = p^n(p-1)$ over $(\mathbb{Z}/p^n\mathbb{Z})^*$, and we have the result.
Is it right or not ? :)
Thank you very much !
The proof looks reasonable, but you should clean up and clarify a few things: what is $q'$ in your first paragraph? Why exactly does $\gcd(m, p(p-1))\neq 1$ imply that $x$ is not a generatator of $(\mathbb{Z}/p^2\mathbb{Z})^\ast$?
Also, I believe in the last paragraph you use $n$ when you mean $n-1$.