Let $p > 2$ be a prime, and let $\zeta = e^{2\pi i/p}.$ Then we know that the minimal polynomial of $\zeta$ is $$f(x) = 1 + x + x^2 + \cdots + x^{p-1} = \prod_{i=1}^{p-1}(x - \zeta^i).$$ Now $\mathrm{Norm}(1-\zeta) = \prod_{i=1}^{p-1}(1 - \zeta^i) = f(1) = p,$ where the norm is the product of the conjugates $\sigma_i(\zeta),$ with the $\sigma_i$ being the $p-1$ embeddings of $\mathbb{Q}(\zeta)$ into $\mathbb{C}$. Furthermore, $\mathrm{Norm}(1-\zeta) = \mathrm{Norm}(\zeta - 1)$ as $p-1$ is even.
Now after this, my notes say that $$f(x+1) = \mathrm{Norm}(x + 1 - \zeta) = \mathrm{Norm}(x - (\zeta -1)) = \text{minimal polynomial of }\zeta - 1.$$ However, I do not really understand how the last equality comes from. Is it not possible that the minimal polynomial of $\zeta - 1$ divides $f(x+1)?$ I'm assuming it uses the previous statement that the norms of $1-\zeta$ and $\zeta - 1$ are the same, but I cannot see how? Can anybody help please?