Say the minimal polynomial of $\alpha$ is some $p(x)\in F[X]$ dividing $X^n-\alpha^n$.
As all roots of $p(x)$ must be roots of $X^n-\alpha^n$, we can write
$$p(x)=\prod_{i\in I} (X-\zeta^i\alpha)$$
where $I$ is some nonempty subset of $\{0,1,2\dots,n-1\}$.
The trick now is just in defining an automorphism on $E[X]$ based on $\sigma$. Specifically, define $\bar \sigma: E[X]\rightarrow E[X]$ so that for any $e(x)=e_0+e_1x+\dots+e_kx^k\in E[X]$, we have $\bar \sigma(e(x))=\sigma(e_0)+\sigma(e_1)x+\dots+\sigma(e_k)x^k$ (I'll leave it to you to check that this an automorphism).
Since $p(x)$ is in $F[X]$, all of $p(x)$'s coefficients are in $F$, so (by the definition of $\bar \sigma$), we ought to have $\bar \sigma(p(x))=p(x)$. This means
$$\bar \sigma\left(\prod_{i\in I} (X-\zeta^i\alpha)\right)=\prod_{i\in I} (X-\zeta^i\alpha)$$
So
$$\prod_{i\in I} \bar\sigma(X-\zeta^i\alpha)=\prod_{i\in I} (X-\zeta^i\alpha)$$
$$\prod_{i\in I} (X-\zeta^i\sigma(\alpha))=\prod_{i\in I} (X-\zeta^i\alpha)$$
$$\prod_{i\in I} (X-\zeta^{i+1}\alpha)=\prod_{i\in I} (X-\zeta^i\alpha)$$
where above we have used the fact that $\zeta\in F$. We get that whenever $i\in I$, $i+1$ (or at least something congruent to $i+1\;\textrm{mod}\; n$) is also in $I$. It follows that $I=\{0,1,2\dots,n-1\}$, so that $p(x)=X^n-\alpha^n$, as desired.
Since, $[F(\alpha):F]=\textrm{deg}(p(x))$ and $[E:F]=n$, it follows directly that $E=F(\alpha)$.
(feel free to comment or edit for any corrections or suggestions)
Best Answer
fresh mans dream: for an integral domain of characteristic $p$ we have $(x-y)^p=x^p-y^p$, in particular $x^{p}-1$ has only $x-1$ as linear factors, which means that there is no $p$-root of unity except for $1$.