I am reading Keith Conrad's notes here on Totally Ramified Primes and Eisenstein Polynomials. I am trying to understand the proof of Lemma 3.1 on page 4 of his notes which I will reproduce here:
Lemma 3.1. Let $K=\Bbb{Q}$ be a number field with degree $n$. Assume $K = \Bbb{Q}(\alpha)$, where $\alpha \in \mathcal{O}_K$ and its minimal polynomial over $\Bbb{Q}$ is Eisenstein at $p$. For $a_0,a_1,\ldots, a_{n-1} \in \Bbb{Z}$, if $a_0 + a_1\alpha + + a_{n-1}\alpha^{n-1} \equiv 0 \pmod{p\mathcal{O}_K}$ then $a_i \equiv 0 \pmod{p\Bbb{Z}}$ for all i.
The proof is as follows. We multiply the expression above by $\alpha^{n-1}$. Since $\alpha$ is Eisenstein at $p$ this eliminates everything except $a_0\alpha^{n-1}\equiv 0 \pmod{p\mathcal{O}_K}.$ Now KCd then says that when we apply $N_{K/\Bbb{Q}}(-)$ we get that
$a_0^n\alpha^{n-1}\equiv 0 \pmod{p^n\Bbb{Z}}.$
My question (1) is: Why is the congruence mod $p^n$ and not mod $p$ ? Is it because $p^n | a_0^n$?
He then proceeds to say:
The norm of is, up to sign, the constant term of its characteristic polynomial for $K=\Bbb{Q}.$ Since $\alpha$ generates $K=\Bbb{Q}$, its characteristic polynomial is its minimal polynomial, which is Eisenstein. Therefore $N_{K/\Bbb{Q}}(\alpha)$ is divisible by $p$ exactly once, so the above congruence modulo $p^n$ implies $p|a_0^n$, so $p|a_0$.
My question (2) is: I get why the norm is divisible by $p$ exactly once, but what does he mean "the above congruence mod $p^n$ implies $p|a_0^n$? This is where I am confused because isn't the congruence $a_0^n\alpha^{n-1}\equiv 0 \pmod{p^n\Bbb{Z}}.$
taken mod $p^n$? Also, how did he deduce that $p|a_0^n$?
Thanks.