$(1-\zeta_m)$ is a unit in $\mathbb{Z}[\zeta_m]$ if m contains at least two prime factors We know that for $m=p^r, 1-\zeta_m$ is a prime.Now suppose that m has at least 2 distinct primes appearing in its prime factorization,we need to show that $1-\zeta_m$ is a unit in its ring of integers $\mathcal{O}_{\mathbb{Q}(\zeta_m)}=\mathbb{Z}[\zeta_m]$. 
I tried proving that $N_{\mathbb{Q}}^{\mathbb{Q}(\zeta_m)} (1-\zeta_m)=\pm 1$ but got stuck in finding norm of $\zeta_m$. Some hint would be nice.

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.
