Showing the lemma $\operatorname{ord}_p(1+ζ_p)=0$ if $p>2$ Just to give some background regarding my motivation, I'm trying to prove a lemma to help me solve How do we prove p-order of $g_k$ is $\frac {k} {p-1}$?
Let $Z_p$ denote the p-adic integers, and let us adjoin a primitive p-th root of unity $ζ_p$. Assume $p>2\DeclareMathOperator{\ord}{ord}$.
I want to show that $\ord_p(1-ζ_p)=\ord_p(1-ζ_p^2)$, but I see that it is enough to prove $\ord_p( \frac {1-ζ_p^2} {1-ζ_p})=0$ or, equivalently, $\ord_p(1+ζ_p)=0$.
I have tried applying the properties from here. (Even though they are for rational numbers, I'm going to assume for now that they hold in $Z_p(ζ_p)$ also. If not, please correct me.)
I see that, if $\ord_p(ζ_p) \neq \ord_p(1)=0$, we will have $\ord_p(1+ζ_p)=inf \{\ord_p(1), \ord_p(ζ_p)\} \leq 0$ and we will get a contradiction if $<0$
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But if instead $\ord_p(ζ_p)=0,$ then I don't know what to do. We will have $\ord_p(1+ζ_p) \geq \inf \{\ord_p(1), \ord_p(ζ_p)\}=0$, but I don't see how $>0$ would give us a contradiction.

Let $$(x-1)^p-1 = f(x)g(x) \in \mathbf{Z}_p[x]$$ where $f(x)$ is the minimal polynomial of $1+\zeta_p$ so $$f(x) = \prod_{\alpha \in Gal(\overline{\mathbf{Q}_p}/\mathbf{Q}_p)\cdot (1+\zeta_p)} (x-\alpha)$$
by definition $|.|_p$ is $Gal(\overline{\mathbf{Q}_p}/\mathbf{Q}_p)$ invariant and$$|1+\zeta_p|_p=|f(0)|_p^{1/\deg(f)}$$
and since $g(0) \in \mathbf{Z}_p$
$$1 \ge |f(0)|_p \ge |f(0)g(0)|_p = |(-1)^p-1|_p = 1$$
whence $|f(0)|_p= 1$ and $|1+\zeta_p|_p = 1$.
