Prove that if $p$ is a prime and $p\ne 2$, then $(\Bbb Z/p^n \Bbb Z)^*$ is cyclic I aim to prove that if $p$ is a prime and $p\ne 2$, then $(\Bbb Z/p^n \Bbb Z)^*$ is cyclic.
It is not obvious, since if $n\ne 1$ then we haven't got a field. May I please ask for a proof? Or any reference would be appreciate.
I can find a quick example that if we take $p=2,n=3$, then $(\Bbb Z/p^n \Bbb Z)^*$ is the Klein four group and thus not cyclic. But may I also ask what is wrong for $p=2$?
Thanks in advance!

Show that $1+p$ has order $p^{n-1}$, therefore $2(1+p)$ has the required order because $2$ has order and divisible by $p-1$, $2^{p-1} \equiv 1 \mod{p} $ (Fermat's little theorem).
to prove the first part use Prove that $1+p$ is an element of order $p^{n-1}$ in $(\Bbb Z/p^n\Bbb Z)^\times$ using the binomial theorem
$|(\mathbb{Z}/p^n\mathbb{Z})^{*}| = (p-1)p^{n-1}$. We have produced an element of the right order..
Appendix: As rightly pointed about by the commentator/questioner propositionx , $2$ is an element of order $(p-1)p^{d}$ (and not p-1 as mistakenly suggested in previous edition), for some $d \in \mathbb{N}$... 
Since $p$ is an odd prime  $\Rightarrow 2^{p-1} \equiv 1 \mod{p}$, which in turn $\Rightarrow 2^{p^k} \not\equiv 1 \mod{p},\ \forall \ k \in \mathbb{N} $. 
Thus $2(1+p)$ must have order exactly equal to $p^{n-1}(p-1)$.
