Show that $U(\mathbb Z_{p^n})$ is cyclic by considering the order of $1+p$ Show that $U(\mathbb Z_{p^n})$ (the group of units) is cyclic for $p$ an odd prime, $n \in \mathbb N$. We are given a hint to consider the order of $1+p$ in this group. I have no idea how this leads to a proof. For example, the order of $4$ in $\mathbb Z_9$ is $3$, so it's clearly not a generator. 
I'm familiar with the proof that uses induction and the lifting lemma, but I don't think that's what we are supposed to use here.
 A: The point is that $U(\mathbb{Z}_{p^n})$ is an abelian group having order $(p-1)p^{n-1}$, so if we find an element having order a multiple of $(p-1)$ (for instance, the generator of $U(\mathbb{Z}_p)$) and an element having order a multiple of $p^{n-1}$, we are done.
Now, we obviously have $(p+1)^k\equiv 1\pmod{p}$ for every $k$, but not every $k$ gives $(p+1)^k\equiv 1\pmod{p^2}$. By the binomial theorem
$$ (p+1)^k = 1+pk+Cp^2, $$
hence in order to have $(p+1)^k\equiv 1\pmod{p^2}$ we must have $k\equiv 0\pmod{p}$. If $n=2$, we proved that the order of $(p+1)$ in $U(\mathbb{Z}_{p^2})$ is a multiple of $p$. If $n>2$, we have that $(1+p)^{pk}\equiv 1\pmod{p^2}$ holds for every $k$, but not every $k$ gives $(1+p)^{pk}\equiv 1\pmod{p^3}$. By the binomial theorem:
$$ (p+1)^{pk}=1+p^2 k+Cp^3, $$
etc. Induction proves that the order of $(p+1)$ in $U(\mathbb{Z}_{p^n})$ is a multiple of $p^{n-1}$, and since we have an element with order $K(p-1)$, $U(\mathbb{Z}_{p^n})$ is a cyclic group. Pretty much the same as the other proof.
