Why $(1-\zeta)$ unit where $\zeta$ is a primitive nth and n divisible by two primes

Solution 1:

Hurkyl's advice in the comments is sensible.

Here is a more theoretical way to think about it; I've never read Lang's book, so I don't know how well it fits with the material of the chapter (but it is a standard argument in number theory):

Write $n = p^k m$ with $p \not\mid m$. Note $(1-\zeta)^{p^k} \equiv 1 - \zeta^{p^k} \bmod p.$

Now $\zeta^{p^k}$ is a primitive $m$th root of $1$, where $p \not \mid m$.

Assuming $m \neq 1$, can you use this to prove that $1 - \zeta^{p^k}$ is a unit mod $p$? And hence that $1 - \zeta$ is a unit mod $p$?

Now find another prime $q$ so that $1 - \zeta$ is also a unit mod $q$.
Once you've done this, you're done. Do you see why?