Let $K$ be a finite extension of $\mathbb{Q}_p$ with residue degree $f$ and ramification degree $e$. Let $R \subset K$ be its ring of integers, let $\mathfrak{p}$ be the prime of $R$ lying above $p$, and let $N_{R/\mathbb{Z}_p}$ be the norm map from $R$ to $\mathbb{Z}_p$.
Question: Is the mod-$\mathfrak{p}$ reduction of $N_{R/\mathbb{Z}_p}$ equal to the $e^{\text{th}}$-power of the norm map $N_{\mathbb{F}_{p^f}/\mathbb{F}_p}$ from $\mathbb{F}_{p^f}$ to $\mathbb{F}_p$?
My attempt at an answer: I think the answer is yes. First, if $e = 1$ (the unramified case), then the answer is yes because the Galois group of $K/\mathbb{Q}_p$ is isomorphic to the Galois group of $\mathbb{F}_{p^f}/\mathbb{F}_p$. Next, if $e > 1$, let $L$ be the maximal unramified extension of $\mathbb{Q}_p$ inside $K$, and let $S$ be the ring of integers of $L$. Then for any $r \in R$, if $\overline{r} \in \mathbb{F}_{p^f}$ denotes the mod-$\mathfrak{p}$ reduction of $r$, there exists $s \in S$ such that the mod-$\mathfrak{p}$ reduction of $s$ is also equal to $\overline{r}$. Because $N_{R/\mathbb{Z}_p}$ is some polynomial function with coefficients in $\mathbb{Z}_p$, the mod-$\mathfrak{p}$ reduction of $N_{R/\mathbb{Z}_p}(r)$ is equal to the mod-$\mathfrak{p}$ reduction of $N_{R/\mathbb{Z}_p}(s)$. But we have that $N_{R/\mathbb{Z}_p}(s) = \big(N_{S/\mathbb{Z}_p}(s)\big)^e$, so the mod-$\mathfrak{p}$ reduction of $N_{R/\mathbb{Z}_p}(s)$ is equal to $\big(N_{\mathbb{F}_{p^f}/\mathbb{F}_p}(\overline{r})\big)^e$, as desired.
It seems to me like the above is a complete proof, but I could not find a reference for this fact anywhere, so I'm wondering if I've missed something.
$L=Q_p(\zeta_{p^f-1})$. Let $N$ be the Galois closure of $K/Q_p$. The ring of integers are $O_L,O_K,O_N$. The maximal ideals are $(\pi_L),(\pi_K),(\pi_N)$.
For $a\in O_K,\not \in (\pi_K)$ there is $m$ such that $a\equiv \zeta_{p^f-1}^m \bmod \pi_K$.
For $\sigma\in Hom_L(K,N)$, $\sigma$ sends $(\pi_K)$ to $(\pi_{\sigma(K)})\subset (\pi_N)$. Thus $$\sigma(a) \equiv \sigma(\zeta_{p^f-1}^m)=\zeta_{p^f-1}^m\equiv a \bmod \pi_N$$
Let $g_1,\ldots,g_f$ be some extensions to $Hom_{Q_p}(N,N)$ of the elements of $Hom_{Q_p}(L,N)$.
$g_j$ sends $(\pi_N)$ to $(\pi_N)$. Whence
$$g_j(\sigma(a)) \equiv g_j(a)\bmod \pi_N$$
$$N_{K/Q_p}(a)=\prod_{j=1}^f \prod_{\sigma\in Hom_L(K,N)} g_j(\sigma(a))\equiv \prod_{j=1}^f \prod_{\sigma\in Hom_L(K,N)} g_j(a) \equiv N_{L/Q_p}(a)^e\bmod \pi_N $$ Which implies that $$N_{K/Q_p}(a)\equiv N_{L/Q_p}(a)^e \bmod p$$