$(1-\zeta_m)$ is a unit in $\mathbb{Z}[\zeta_m]$ if m contains at least two prime factors

Write $1+x+x^2+...+x^{n-1} = \prod_{j=1}^{n-1} (x-\zeta_n^j)$ and put x=1 to get $n = \prod_{j=1}^{n-1} (1-\zeta_n^j)$. If $p^a||n$, then running $j$ through multiples of $n/p^a$, we see that the product contains $p^a = \prod_{j=1}^{p^a-1} (x-\zeta_{p^a}^j)$. Remove all such factors and get $1 = \prod (1 - \zeta_n^j)$ with the product over the $j$ which are not prime powers. By your assumption, $n$ is not a prime power, so you will have $(1-\zeta_n)$ in this product, therefore it is a unit.

Edit: If you extend this proof and write the product properly, you will actually see that the norm is +1.

If $m$ is not a prime power, then there exist $a$, $b \in \mathbb{Z}$ such that $ab = m$ and $\gcd(a,b) = 1$. Set $\zeta_{a} := \zeta_{m}^{b}$ and $\zeta_{b} := \zeta_{m}^{a}$; these are primitive $a$th and $b$th roots of unity, respectively. Thus, we have the standard factorizations $$a = \prod_{0\neq j \in \mathbb{Z}/a\mathbb{Z}} (1-\zeta_{a}^{j})$$ $$b = \prod_{0\neq j \in \mathbb{Z}/b\mathbb{Z}} (1-\zeta_{b}^{j})$$ In particular, we see that in $\mathbb{Z}[\zeta_m]$, we have $1-\zeta_m$ divides $1-\zeta_{a}$ divides $a$, and similarly $1-\zeta_m$ divides $b$. On the other hand, since $\gcd(a,b) = 1$ in $\mathbb{Z}$, Bezout's lemma gives integers $x$ and $y$ such that $xa + yb = 1$. It follows that $1-\zeta_m$ divides $1$ in $\mathbb{Z}[\zeta_m]$, i.e. $1-\zeta_m$ is a unit.