Let $\mathbb{Z}_p$ be the ring of $p$-adic numbers, $\Phi _p(x)$ the cyclotomic polynomial and $\zeta$ be a $p$-th root of unity. So $\zeta \not \in \mathbb{Z}_p$ and we get $$ \mathbb{Z}_p[X]/\langle\Phi_p(x)\rangle \cong \mathbb{Z}_p[\zeta] = \mathbb{Z} \oplus \zeta \mathbb{Z} \oplus \dots \oplus \zeta^{p-2} \mathbb{Z} $$ a free $\mathbb{Z}_p$-module of rank $p-1$.
Let $\pi := \zeta-1$ and I want to know how to see that $\mathbb{Z}_p[\zeta]$ is a local ring with maximal ideal $\pi \mathbb{Z}_p[\zeta]$.
My idea is to use the obviously fact that $\Phi _p(\pi +1) =0$ and the extension formula $ \Phi_p(x + 1) = p + \binom{p}{2}x + \binom{p}{3}x^2 + \dots + \binom{p}{p - 1} x^{p - 2} + x^{p - 1} $.
From this I can conclude that we have following inclusions: $p \mathbb{Z}_p[\zeta] \subset \pi \mathbb{Z}_p[\zeta]$ and $\pi^{p-1} \mathbb{Z}_p[\zeta] \subset p \mathbb{Z}_p[\zeta]$
It's obviosly that it would be enough to show that $\mathbb{Z}_p[\zeta] / \pi\mathbb{Z}_p[\zeta] \cong \mathbb{F}_p$ and that every maximal ideal $M \subset \mathbb{Z}_p[\zeta]$ contains $\pi$ but unfortunately I haven't a idea how to realize this two steps using the information above...
$\Phi_p(\pi + 1) = 0$ tells you that
$$ \Phi_p(x + 1) = p + \binom{p}{2}x + \binom{p}{3}x^2 + \dots + \binom{p}{p - 1} x^{p - 2} + x^{p - 1} $$
is the minimal polynomial of $\pi$. What does that imply about $(\pi\mathbf{Z}_p[\zeta]) \cap\mathbf{Z}_p$? Which ideal is that?