The minimal polynomial of a primitive $p^{m}$-th root of unity over $\mathbb{Q}_p$

Proposition 7.13 of Neukirch's ANT states that for a primitive $p^{m}$-th root of unity $\zeta$ ($p$ prime) the extension $\mathbb{Q}_{p}(\zeta)/\mathbb{Q}_{p}$ is totally ramified of degree $(p-1)p^{m-1}$. In the proof he shows that the polynomial $\phi(X)=X^{(p-1)p^{m-1}}+X^{(p-2)p^{m-1}}+\cdots+1$ is the minimal polynomial of $\zeta$ over $\mathbb{Q}_p$ .

I can't see how to obtain the congruence $\phi(X)=(X^{p^m}-1)/(X^{p^{m-1}}-1)\equiv(X-1)^{p^{m-1}(p-1)}$ mod $p$ as claimed in the book. Here, I take it that mod $p$ means mod $p\mathbb{Z}$ and not mod $p\mathbb{Z}_{p}$ ?

Also, does Eisenstein's irreducibility criterion extend to polynomials in $\mathbb{Q}_{p}[X]$ ?

Any help would be very much appreciated.

In $\mathbf{F}_p[X]$ (you are looking at the polynomial $p\mathbf{Z}$, so as an element of $\mathbf{F}_p[X]$), because you're in characteristic $p$, $X^{p^m}-1=(X-1)^{p^m}$, and similarly $X^{p^{m-1}}-1=(X-1)^{p^{m-1}}$, so the quotient, modulo $p$, is $(X-1)^{p^m-p^{m-1}}=(X-1)^{p^{m-1}(p-1)}$. This is the congruence

$\phi(X)=(X^{p^m}-1)/(X^{p^{m-1}}-1)\equiv (X-1)^{p^m(p-1)}\pmod{p}$.

Eisenstein's criterion holds for any UFD $R$ and its field of fractions. In particular it holds for $\mathbf{Z}_p$ and $\mathbf{Q}_p$ (using the prime ideal $(p)$ of $\mathbf{Z}_p$).

EDIT: To answer the question posed in the coments, in $\mathbf{F}_p[X]$, $X^{p^m}-1=(X-1)^{p^m}$ and $X^{p^{m-1}}-1=(X-1)^{p^{m-1}}$. Since the latter clearly divides the former (they are both powers of the same polynomial and the former is a larger power) the quotient is an element of $\mathbf{F}_p[X]$, which can be checked directly to be the image (i.e. reduction) of $\phi(X)$ in $\mathbf{F}_p[X]$, simply because, in $\mathbf{Z}[X]$, $\phi(X)=(X^{p^m}-1)/(X^{p^{m-1}}-1)$ (this equality is written in the OP's question so I assume it is understood).