I think I heard somewhere long ago that $\mathbb{Z}/p^k\mathbb{Z}$'s unit group is cyclic if $p$ is odd and $C_2\times C_{2^{k-2}}$ if $p=2$. I remember trying to prove it and finding it surprisingly difficult.
This morning I think I realized you can prove this using the $p$-adic exponential and log maps. Proof sketch below. In a way, this result seems like the $p$-adics would be its natural home, because it's about $\mathbb{Z}$ mod $p^k$ for arbitrary $k$. However, it also seems like exactly the sort of completely classical result that Gauss should have a proof of somewhere. So, my question:
How is this usually proven? And, is there a classical, Gauss-typed or at least 19th-century-typed proof?
Proof sketch for $p$-adic proof: if $p$ is odd, then $\exp_p$ is a continuous group isomorphism from $(p\mathbb{Z}_p)^+$ to $(1+p\mathbb{Z}_p)^\times$. Since $p\mathbb{Z}\subset p\mathbb{Z}_p$ is cyclic and dense, this means that the $1$-units of $\mathbb{Z}_p$ have a dense cyclic subgroup. Reducing mod $p^k$ means that the $1$-units of $\mathbb{Z}/p^k\mathbb{Z}$ are cyclic, since the group is discrete so a dense subset is everything. Now I just have to show that some generating $1$-unit is a $(p-1)$th power in $\mathbb{Z}/p^k\mathbb{Z}$, and this I can do. If $p=2$, then I don't have all the details worked out, but $\log_p$ is $2$-to-$1$ from $1+2\mathbb{Z}_2$ to $2\mathbb{Z}_2$; and I think the preimage of $2\mathbb{Z}$ is dense, whereupon since it is $\{\pm 1\} \times$ cyclic, generated by a lift of $2$, the same argument works (without the shenanigans about the $1$-units, since every unit in $\mathbb{Z}/2^k\mathbb{Z}$ is a $1$-unit).
