Let $p$ be a prime, $n$ a positive integer and $q=p^n$ a $p$- power. Let $w,v$ is a root of the $q,p$-th cyclotomic polynomial $Q_q,Q_p$, respectively. Let $f$ be a polynomial with coefficient in $\mathbb Z$.
Prove or disprove: "$p^a\mid f(v) \implies p^a\mid f(w)$" in $\mathbb Z[w]$.
Edit: This question was an "iff" question, now it is "implies" question.
Moreover: $(p)=(1-v)^{p-1}$ in $\mathbb Z[v]$ and $(p)=(1-v)^{p^{n-1}(p-1)}$ in $\mathbb Z[w]$. That is, $p$ is totally ramified in both extension.
If $i$ is coprime to $p$ this follows immediately from the fact that there is an automorphism $\sigma$ of the field $\Bbb{Q}(w)$ that maps $w\mapsto w^i$. Because $f\in\Bbb{Z}[x]$, we have $$ f(w^i)=f(\sigma(w))=\sigma(f(w)). $$ So if $f(w)=p^az$ for some number $z\in\Bbb{Z}[w]$ then $f(w^i)=p^a\sigma(z)$. Apply the inverse of $\sigma$ to get the other implication.
If $p\mid i$, then I think the claim is false. Let $f=Q_q$. Then $f(w)=0$ and thus divisible by $p^a$ for all $a$. OTOH $f(w^p)=p$ is divisible by $p^a$ only when $a=0,1$.