Let $K$ be a cyclic extension of a field $F$, with Galois group $G=\langle \sigma \rangle$ and assume that $\operatorname{char}F=p$ and that $[K:F]=p^{m-1}$ for some $m>1$. Let $\beta$ be an element in $K$ such that $\mathrm{Tr}_k^K(\beta)=1$.
(a) Show that there exists an element $\alpha \in K$ such that $\sigma\alpha-\alpha=\beta^p-\beta$.
(b) Prove that the polynomial $X^p-X-\alpha$ is irreducible in $K[X]$.
(c) If $\theta$ is a root of this polynomial, prove that $F(\theta)$ is a Galois, cyclic extension of degree $p^m$ of $F$, and that ist Galois group is generated by an extension $\sigma^\ast$ of $\sigma$ such that $\sigma^\ast(\theta)=\theta+\beta$.
This problem can be found on Serge Lang's Algebra 3rd Edition and in this question post. However the answer there doesn't work to me and I am trying a different approach so please do not close as duplicate instantly.
I have solved part (a) and (b) but stuck at (c). Here's my attempt so far.
By Artin-Schreier theorem, $K(\theta)$ is a cyclic extension of $K$ with degree $p$. The Galois group $G(K(\theta)/K)$ is a cyclic group generated by $\lambda:\theta \mapsto \theta+1$, as $\theta+i$ is a root of $X^p-X-\alpha$ for $i=0,1,\dots,p-1$.
Therefore $[K(\theta):F]=p^m$. On the other hand we have $F(\theta) \subset K(\theta)$ therefore it is desirable to show that the degree of $\theta$ over $F$ is $p^m$, which will imply that $F(\theta)=K(\theta)$.
For the relation $\sigma^\ast(\theta)=\theta+\beta$, I suspect we can deduce it through the fact $\mathrm{Tr}^{F(\theta)}_{F}(\beta)=\mathrm{Tr}^K_F \circ \mathrm{Tr}^{F(\theta)}_{K}(\beta)=\mathrm{Tr}_F^K(0)=0$, and apply Hilbert's theorem 90 to give rise to an element $\zeta \in F(\theta)$ such that $\sigma^\ast(\zeta)-\zeta=\beta$, and find out that $\zeta=\theta$ with no surprise.
I also suspect that in fact $K=F(\beta)$ because $G(K(\theta)/K)$ is generated by $\theta \mapsto \theta+1$ and $G(F(\theta)/F)=G(K(\theta)/F)$ is generated by $\theta \mapsto \theta+\beta$.
Anyway, my question (hence the title) is, how to determine the degree of $\theta$? Is it possible to find the minimal polynomial of $\theta$, which should also have $\theta+\beta$ as a root? I suppose by having 1 done (where 3 may also play a role), we can show that $F(\theta)/F$ is indeed Galois and cyclic and determine $\sigma^\ast$ (perhaps through 2).