Let $p$ be a prime. Let $n$ and $k$ be positive integers such that $k$ divides $pn$ but not $n$ (that is, $k$ is a divisor of $pn$ having $p$-adic order that is one greater than the $p$-adic order of $n$).
Let $E = GF(p^{pn})$, let $F = GF(p^n)$, and let $K = GF(p^k)$.
The Problem:
I want to show that the image of $K$ under the mapping $x \mapsto N_{E/F}(x)$ is equal to $GF(p^{k/p})$.
What is the best way to do this?
At this point this is more of a comment than an answer, I'm more or less "thinking out loud" here. Downvote if you wish.
I'm going to deal first with $p=2$. Let $N_K$ denote the image of $K$ in $F$ under the mapping $x \mapsto N_{E/F}(x)$.
Clearly we have $0 \in N_K$, and furthermore the only element that maps to $0$ is $0$. It is also clear that $1 \in N_K$ and that $N_K$ is closed under multiplication, since the norm is multiplicative. For closure under addition I'm not so sure... perhaps there is something to be done by representing the elements of $N_K$ as polynomials over $GF(2)$?
For $x \in K$ we have $N_{E/F}(x) = x^{2^n+1}$. It can be shown that the size of the image of $K^*$ under the mapping $x \mapsto x^{2^n+1}$ is $$\frac{2^k-1}{\gcd(2^k-1,2^n+1)} = \frac{2^k-1}{2^{\gcd(k,n)}+1} = \frac{2^k-1}{2^{k/2}+1} = 2^{k/2}-1.$$ Therefore $N_K$ is a subset of $F$ that has size $2^{k/2}$, contains $0$ and $1$, and is closed under multiplication.
Is this enough for this case?