Prove $L = F(β)$, where $β^p = F$ via Hilbert's Theorem 90 Let $F$ be a field that contains a primitive $p$-th root of unity, where $p$ is a prime. I wish to prove that if $L$ is Galois over $F$ and $[L : F] = p$, then $L = F(β)$, where $β^p = F$. Does anyone know of a proof via Hilbert's Theorem 90? 
That is, making use of the fact that: Given $Gal(L|F)$ is cyclic of order $n$ and $σ$ is a generator of $G$, suppose $δ_0 = α, δ_1 = ασ(α), δ_2 = ασ(α)σ^2(α), · · · ,δ_{n−1} = ασ(α)· · · σ^{n−1}(α) =$ 'norm' $N(α) = 1$ and $γ ∈ L$ is such that
$β = δ_0γ + δ_1σ(γ) + · · · + δ_{n−1}σ
^{n−2}(γ) + σ^{n−1}(γ) ≠ 0$. Then we have $α = βσ(β)^{−1}$

Let $\zeta$ be a primitive $p$-th root of unity in $F$. Then $N_{L/F}(\zeta)=\zeta^p=1$. By Hilbert 90, there is $\alpha\in L^*$ with
$\sigma(\alpha)/\alpha=\zeta$, that
is
$$\sigma(\alpha)=\zeta\alpha.$$
But then
$$\sigma(\alpha^p)=\sigma(\alpha)^p=\zeta^p\alpha^p$$
so that $\alpha^p=a\in F$. But $\sigma(\alpha)\ne\alpha$ so $\alpha\notin F$.
Thus $L=F(\alpha)$ with $\alpha^p\in F$.
