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

From Chapter VII of Lang’s Algebra.

The question asks if $n\geq 6$ and $n$ is divisible by at least two primes, show that $1-\zeta$ is a unit in the ring $\mathbb{Z}[\zeta]$

I am having a hard time understanding why this is true. This is in the integral dependence chapter, but that has not given me any inspiration. I have also tried using cyclotonic polynomial to no avail

Thanks for any direction.

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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?