Q:

Field Norm Surjective for Finite Extensions of $\mathbb{F}_{p^k}$

I'd rather not have the answer, because I feel like this should be a relatively easy question, and I'm just missing some key step, but could anyone give me a hint on showing that the norm (defined as $N(a)=\det(L_a)$) where $L_a$ is the linear transformation given by multiplication by $a$ is surjective in the case of a finite extension of a finite field?
I've been looking at the fact that $N(x)=x^n$ for any $x$ in the base field, where $n$ is the degree of the field extension.  But this doesn't necessarily give me back every element in the base field, for instance $\mathbb{F}_3(\sqrt{2})$, where we have to apply norm to $\sqrt{2}$ to get $2$ back.  Is there a basic fact about finite fields maybe that I'm missing, or something more clever regarding a field extension?
Thanks!

A:

Two facts:

The norm is multiplicative: $N(ab) = N(a)N(b)$.
The nonzero elements of a finite field form a cyclic group (under multiplication).

So, if you can figure out what a generator of the multiplicative field of $\mathbb{F}_{p^k}$ maps to...
Added. Your comment below suggests you are a bit confused. So let me set things up a bit.
Given a field extension $F\subseteq K$ of finite degree, the norm from $K$ to $F$, $N_{K/F}\colon K\to F$ is the map that sends $N(a)$ to the determinant of $L_a\colon K\to K$, the linear transformation from $K$ to $K$ given by multiplication by $a$, considering $K$ as a vector space over $F$. The map is multiplicative, and always takes values in $F$. "Surjectivity" here would refer to surjectivity as a map $N_{K/F}\colon K\to F$.
You are considering $F=\mathbb{F}_{p^k}$, and $K=\mathbb{F}_{p^{kn}}$ for some positive integers $k$ and $n$ (remember that $\mathbb{F}_{p^a}$ is an extension of $\mathbb{F}_{p^b}$ if and only if $b|a$).
Since the multiplicative group of $K$ is cyclic, it is generated by some $a$; so the non-zero part of the image of $N$ is generated by $N(a)$, hence you only need to figure out what $N(a)$ is.
Because the powers of $a$ give all nonzero elements of $K$, then $K=\mathbb{F}_{p^k}(a)$; so $\{1,a,a^2,\ldots,a^{n-1}\}$ is a basis for $K$.
It is pretty easy to figure out what the matrix of $L_a$ is with respect to this basis (it will depend on the minimal polynomial of $a$, though). Then you want to argue that the determinant of this matrix is necessarily a generator of the multiplicative group of $\mathbb{F}_{p^k}$. 
For example, with $\mathbb{F}_3(\sqrt{2})$, the multiplicative group is generated by $1+\alpha$, where $\alpha=\sqrt{2}$, since:
$$\begin{align*}
(1+\alpha)^2 &= 1+2\alpha+\alpha^2 = 3+2\alpha = 2\alpha;\\
2\alpha(1+\alpha) &= 2\alpha+4 = 1+2\alpha;\\
(1+2\alpha)(1+\alpha) &= 2;\\
2(1+\alpha) &= 2+2\alpha;\\
(2+2\alpha)(1+\alpha) &= \alpha;\\
\alpha(1+\alpha) &= 2+\alpha;\\
(2+\alpha)(1+\alpha) &= 1.
\end{align*}$$
Note that $\{1,1+\alpha\}$ is a basis for $\mathbb{F}_3(\sqrt{2})$ over $\mathbb{F}_3$. If we let $a=1+\alpha$, then $L_a$ has matrix, relative to this basis, equal to
$$\left(\begin{array}{cc}
0 & 1\\
1 & 2
\end{array}\right)$$
(since $(1+\alpha)^2 = 2\alpha = 1 + 2(1+\alpha)$). So the determinant of this matrix is $-1 = 2$, hence $N(1+\alpha) = 2$, which happens to be a generator of $\mathbb{F}_3^{\times}$. That means that the image of $N$ consists of $0$ plus the subgroup of $\mathbb{F}_3^{\times}$ generated by $N(1+\alpha)=2$, which is all of $\mathbb{F}_3$. 
Note that even though $K=\mathbb{F}_3(\sqrt{2})$, $\sqrt{2}$ does not generate the multiplicative group of nonzero elements of $K$; we needed to take a different element.

A:

Remark: I replied too soon and provided an answer for the trace of a field extension instead. 
I won't delete it because it's somewhat interesting, even though it's useless.
I have included but different calculation for the norm just for completion of my reply.
$$ N\colon \mathbb F_{q^\ell}^*\to \mathbb F_{q^\ell}^* \colon x\mapsto xx^q x^{q^2}\cdots x^{q^{\ell-1}}  = x^{\tfrac{q^\ell-1}{q-1}} $$
Notice that $N$ is a morphism of cyclic groups and that in fact $\mathrm{im} N \subseteq \mathbb F_q^*$. 
Now it's well known (or else easy to show) that if 
$$ \varphi \colon C\to C \colon x\mapsto x^m $$
is a morphism of cyclic groups, then $|\mathrm{im} \varphi| = |C|/(m,|C|)$. 
In this case $|\mathrm{im} N |= \frac{q^\ell-1}{(q^\ell-1,\frac{q^\ell-1}{q-1})} = q-1$ and indeed $N$ is onto $\mathbb F_q^*$.

Original reply:
I'm not sure where exactly you are stuck, so I wouldn't know what to hint, but here is how I usually deal with such a problem.
Let $\mathbb F_{q^\ell}/\mathbb F_q$ be a field extension, with associated trace
$$ T\colon \mathbb F_{q^\ell}\to \mathbb F_q \colon x\mapsto x+x^q + x^{q^2} + \dots + x^{q^{\ell-1}} $$
Notice that $T$ is a morphism of additive groups. (And note that indeed $T(x)^q = T(x)$ for all $x$ so it goes to $\mathbb F_q$ indeed).
Now first of all, all elements in $\ker T$ are solutions to an equation of degree $T^{\ell-1}$ therefore $|\ker T|\le q^{\ell-1}$. On the other hand $|\mathrm{im}\, N| = q^\ell/ |\ker N| \leq q$ therefore $|\ker T| \geq q^{\ell-1}$. Therefore $|\ker T|=q^{\ell-1}$ and $|\mathrm{im} T|=q$ and $T$ is a surjection.
-- Edit. I should clarify that the definition of "Trace" I use may seem different from yours; fortunately Arturo Magidin --as always-- provides an detailed explanation on that.

