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.