Quadratic Extension of Finite field Suppose $K,L$ are finite fields with $|K|=p^n$ and the $L$ is a quadratic extension over $K$, i.e. $|L| = p^{2n}$. I am trying to show that for any element in the extension $\alpha\in L$ that $\alpha^{p^n+1} \in K$ and moreover that every element in $K$ is of the can be represented as $\alpha^{p^n+1}$ for $\alpha\in L$ . Further, I want to show that if an element in $\beta \in K$ is a generator for $K^*$, i.e. has order $p^n-1$, then there is a generator $\alpha\in L$, i.e. of order $(p^n)^2-1$, such that $\alpha^{p^n+1}=\beta$.
I tested this out with concrete examples with $p=2,3$ and $n=1,2$ but I couldn't really gain much insight from that. I am not sure if this needs to be broken into cases or something. I know that from the extension being quadratic every element $\alpha \in L$ satisfies some irreducible quadratic polynomial in $K$. That is $\alpha^2+b\alpha+c=0$ for some $b,c \in K$. I'm not sure if I need to specify whether $p$ must be odd, but the book does not seem to do so. More stuff I tried with the odd assumption was $\alpha^{p^n+1}=(\alpha^2)^{(p^n+1)/2}= (b\alpha+c)^{(p^n+1)/2}$  for some $b,c \in K$. This does not seem to lead anywhere useful though. 
Also, even if I was assume the first part and moreover part, I am still confused by the "further part" . By the moreover part $\beta$ has some representation as $\alpha^{p^n+1}$. We know that $\beta^{p^n-1}=1$ thus $(\alpha^{p^n+1})^{p^n-1}=\alpha^{p^{2n}-1}=1$. Suppose though the order of $\alpha$ was some smaller number $d|p^{2n}-1$, i.e. $\alpha^d=1$. Then $\beta^d=(\alpha^{p^n+1})^d=(\alpha^d)^{p^n+1}=1$. But since $\beta$ is a generator of $K$ the $d$ must be a multiple of $p^n-1$ which does divide $p^{2n}-1$, with remainder $p^n+1$. So it seems that $\alpha$ is allowed to have order $p^n-1$. This would mean though that $\beta=\alpha^{p^n+1}=\alpha^{p^n-1}\alpha^2=\alpha^2$. I am not sure where to go from there though, not even sure if this is the right direction. 
 A: Let $q = p^n$.
The roots of the polynomial $x^q - x$ are precisely the elements of $\mathbb{F}_q$. If $\alpha \in \mathbb{F}_{q^2}$, then $\alpha^{q^2} - \alpha = 0$, and
$$ \left(\alpha^{q+1}\right)^q = \alpha^{q^2 + q} = \alpha^{q^2} \cdot \alpha^q = \alpha \cdot \alpha^q = \alpha^{q+1} $$
Therefore $\alpha^{q+1} \in \mathbb{F}_q$.
In general, given a finite extension $L/K$ of fields, there is the concept of the norm (over $K$) of an element of $L$, which is the product of all of its conjugates, and is an element of $K$. If $K = \mathbb{F}_q$, the conjugates of any element are obtained by repeatedly raising it to the $q$-th power.
In this case, we have a quadratic extension, so
$$N(\alpha) = \alpha^q \cdot \alpha = \alpha^{q+1}$$
which gives another proof that $\alpha^{q+1} \in \mathbb{F}_q$.
To see that every element of $\mathbb{F}_q$ is a norm -- that is, a $(q+1)$-th power of an element in $\mathbb{F}_{q^2}$ -- we can appeal to polynomial equations again. Every $\alpha \in \mathbb{F}_{q^2}$ is a root of some equation
$$ f_\beta(x) = x^{q+1} - \beta = 0 $$
and each such equation has at most $q+1$ roots. $f_0(x)$ has a single root $0$ with multiplicity $q+1$. The remaining $q^2 - 1 = (q-1)(q+1)$ elements of $\mathbb{F}_{q^2}$ are distributed among the polynomials corresponding to remaining $q-1$ values for $\beta$, with each $f_\beta$ having at most $q+1$ of them. A counting argument thus shows that every $f_\beta(x)$ with $\beta \neq 0$ must have exactly $q+1$ distinct roots in $\mathbb{F}_{q^2}$.
We could (as the other answers have shown) argue by using the fact the unit group of $\mathbb{F}_{q^2}$ is cyclic.
Having the idea of the norm, it's easy to generalize to arbitrary finite extensions of $\mathbb{F}_q$: the norm of an element of $\mathbb{F}_{q^n}$ is
$$ N(\alpha) = \alpha \cdot \alpha^q \cdot \ldots \cdot \alpha^{q^{n-1}}
= \alpha^{(q^n-1) / (q-1)}$$
Of course, this again could have been discovered by using the fact the unit group is cyclic.
(For extensions $L/K$ of arbitrary fields, it's not always true that every element of $K$ is a norm)
A: Work in progress
The litmus test for deciding whether an element $\beta$ in an extension field 
$L = \mathbb F_{q^m}$ belongs to the base field $K = \mathbb F_q$ is to check
whether $\beta^q = \beta$ or not. If $\beta^q = \beta$, then $\beta \in K$;
else $\beta \notin K$. In your case, $m = 2$. Given any $\alpha \in L = \mathbb{F}_{q^2}$,
write $\alpha^{q+1} = \beta$, and note that
$$\beta^q = \left(\alpha^{q+1}\right)^q = \alpha^{q^2+q} = \alpha^{q^2}\cdot\alpha^q
= \alpha\cdot\alpha^q = \beta$$
where we have used the litmus test $\alpha^{q^2}\stackrel{?}{=}\alpha$ for membership in $\mathbb F_{q^2}$ to aver that $\alpha^{q^2} = \alpha$.  Applying the litmus test for membership in $\mathbb F_q$ to $\beta = \alpha^{q+1}$, we have that
$\alpha^{q+1} \in \mathbb F_q$ for all $\alpha \in \mathbb F_{q^2}$.  In other words, the function $x \mapsto x^{q+1}$ maps $L=\mathbb F_{q^2}$ to $K = \mathbb F_q.$
Turning to the matter of the orders of the elements, let $\alpha \in L$
denote a generator of $L^*$ so that ${\mathsf{ord}}(\alpha) = q^2-1$.
Then
$${\mathsf{ord}}(\alpha^{q+1}) = \frac{{\mathsf{ord}}(\alpha)}{\gcd({\mathsf{ord}}(\alpha),q+1)}
= \frac{q^2-1}{\gcd(q^2-1,q+1)} = q-1$$
and so whenever $\alpha$ is
a generator of $L^*$, it is mapped by the transformation $x \mapsto x^{q+1}$ onto $\beta=\alpha^{q+1}$ which is a generator of $K^*$: generators of $L^*$  are mapped into generators of $K^*$  Now, since $\beta$ generates $K^*$, every generator $\gamma$ of $K^*$ can be expressed as a power of $\beta$, that is, for each generator $\gamma$ of $K^*$, there exists an integer $k$, relatively prime to $q-1$, such that $\gamma = \beta^k$. Then,
$$\left(\alpha^k\right)^{q+1} = \alpha^{k(q+1)} = \left(\alpha^{q+1}\right)^k
= \beta^k = \gamma.$$ 
Is $\alpha^k$ a generator of $K^*$? Well,
$${\tt{ord}}(\alpha^k) = \frac{{\tt{ord}}(\alpha)}{\gcd({\tt{ord}}(\alpha),k)}
= \frac{q^2-1}{\gcd(q^2-1,k)}$$
Furthermore, we can use the fact that 
$\tt{ord}(\alpha) = q^2-1$ to deduce that for any integer $i$,
$$\left(\alpha^{k+i(q-1)}\right)^{q+1} = \alpha^{k(q+1)+i(q^2-1)} 
= \alpha^{k(q+1)}\cdot\alpha^{i(q^2-1)} = \alpha^{k(q+1)}
= \beta^k = \gamma.$$
Thus, the $q+1$ elements $\alpha^k, \alpha^{k+(q-1)}, \alpha^{k+2(q-1)},
\ldots, \alpha^{k+q(q-1)}$ are the 
$q+1$ roots of $x^{q+1} - \gamma \in \mathbb F_q[x]$.
At least one of these elements must be a generator of $L^*$.
Work in progress
A: HINT: 
The multiplicative group of a finite field is always cyclic. Thus $L^\times$ is cyclic of order $p^{2n}-1=(p^{n}+1)(p^n-1)$ and $K^\times$ is reobtained back as the only cyclic subgroup of order $p^n-1$.
By writing $L^\times=\langle\gamma\rangle$ and writing every element as a power of $\gamma$ all the answers to your questions should be straightforward.
A: I keep your notations K, L, q etc. We have a cyclic extension L/K, and from the theory of finite fields, Gal(L/K) is  generated by the Frobenius automorphism defined by Fr(x) = $x^q$ for any x in L. The norm map N from L* to K* is defined by taking the product of conjugates, i.e. N(x) = x. Fr(x) = $x^(q+1)$. Since $(Fr)^2$ = Id, it is obvious that Fr(N(x)) = N(x), hence N(x) belongs to K, as desired.
Let us determine the image of the norm. Since N is the same as raising to the (q+1)-power in the cyclic group L*, Ker N has order (q+1), hence Im N has order $(q^2 - 1)/q + 1$ = q - 1 = order of K* , i.e. the norm is surjective, as desired.
