Let $K$ be a field of characteristic $p$. $L=K(\alpha)$ with $\alpha^p-\alpha=a\in K$, an extension of order $p$. Show that there does not exist $\beta \in L$ such that $\alpha^{p-1}=\beta ^p-\beta$.
I tried to use Norm and Trace, but have not succeeded.
I'm afraid the claim is still false. Consider the case $K=\Bbb{F}_4=\{0,1,\gamma,\gamma+1\}$ with $\gamma^2=\gamma+1$. Let us pick $a=\gamma$. Let $\alpha$ be a zero of the polynomial $p(x)=x^2+x+\gamma$. Then $\alpha^2$ is a zero of $r(x)=x^2+x+\gamma^2$, because $r(\alpha^2)=p(\alpha)^2=0^2=0$. But the polynomial $$ p(x)r(x)=(x^2+x)^2+(\gamma+\gamma^2)(x^2+x)+\gamma^3=(x^4+x^2)+(x^2+x)+1=x^4+x+1 $$ is known to be irreducible over the prime field, and is thus the minimal polynomial of $\alpha$ over $\Bbb{F}_2$. Because this polynomial has no cubic term, we see that $\mathrm{tr}^L_{\Bbb{F}_2}(\alpha)=0$. Therefore $\alpha=\beta^2-\beta$ for some $\beta\in K=\Bbb{F}_{16}$. Indeed we see that $\beta=\gamma(\alpha+1)$ fits the bill as $$ \begin{aligned} \beta^2-\beta&=\gamma^2\alpha^2+\gamma^2+\gamma\alpha+\gamma\\ &=(\gamma+1)(\alpha+\gamma)+\gamma^2+\gamma\alpha+\gamma\\ &=\alpha. \end{aligned} $$
What I do below is wrong in the sense that it does not answer the question. I only take the trace from $L$ to $K$, but that version of Hilbert 90 uses the absolute trace, i.e. the smaller field should be the prime field. By transitivity of trace we see that the absoluet trace of $\alpha^{p-1}$ is $[K:\Bbb{F}_p]\cdot(-1_K)$, and this may well vanish. In fact that is how I produced the above counterexample. Leaving it here as it may be marginally helpful for tracking the correct version of the question/claim. I do correct the various trace functions
A small step towards a solution explaining why the claim holds, if the field $K$ (and hence also $L$) is finite. The solution relies on the well known additive version of Hilbert's Satz 90 stating that an element $x$ of a finite field $F$ of characteristic $p$ can be written in the form $x=y^p-y$ for some $y\in F$, iff $\mathrm{tr}^F_{\Bbb{F}_p}(x)=0$. For an on-site proof, see this recent answer by yours truly.
The Galois group $G=\mathrm{Gal}(L/K)$ is cyclic of order $p$. It consists of the mappings $\sigma_j$ determined by $\sigma_j(\alpha)=\alpha+j$ for $j=0,1,2,\ldots,p-1$. Therefore
$$
\begin{aligned}
\mathrm{tr}^L_K(\alpha^{p-1})&=\sum_{\sigma\in G}\sigma(\alpha^{p-1})=\sum_{j=0}^{p-1}\sigma_j(\alpha^{p-1})\\
&=\sum_{j=0}^{p-1}(\alpha+j)^{p-1}=\sum_{j=0}^{p-1}\sum_{k=0}^{p-1}\binom{p-1}k \alpha^kj^{p-1-k}\\
&=\sum_{k=0}^{p-1}\alpha^k\binom{p-1}k\left(\sum_{j=0}^{p-1}j^{p-1-k}\right).
\end{aligned}
$$
Here the inner sum of powers of $j$ is easily seen to vanish unless $k=0$, in which case it consists of $p-1$ ones and a single zero. The vanishing part of this claim also follows from the fact that $\mathrm{tr}^L_K$ takes values in the smaller field $K$. Therefore
$$
\mathrm{tr}^L_K(\alpha^{p-1})=\alpha^0\cdot\binom{p-1}0\cdot(p-1)=-1_K.
$$
As this is non-zero, the claim follows from that additive Hilbert 90. At this point we can conclude (using another application of Hilbert 90) that the element $\alpha^{p-1}$ cannot be written in the form $\beta^{|K|}-\beta$ for some $\beta\in L$. As the counterexample above shows, we cannot replace the exponent $|K|$ with $p$ here, unless we know that the degree of the extension $[K:\Bbb{F}_p]$ is coprime to $p$.
