http://www.math.uiuc.edu/~r-ash/Algebra/Chapter6.pdf
In the pdf, corollary $6.4.2$, the extension $E/F_p$ is Galois. Why is it Galois? Is it because $F_p$ is a finite field and hence every polynomial with coefficient in $F_p$is separable which gives us the separability of all minimal polynomial with root $\alpha \in E$ over $F_p$?
Also, why is that $x^p=x$ for all $x \in F_p$? It seems like it got something to do with the polynomial $X^{p^{n}}-X$ but I don't know the relation.
Observation 1 Note that if $|E| = p^n$ (this happens when $[E:F_p] = n$ $(*)$), then $g^{p^n - 1} = 1$ for each $g\in E\setminus\{0\}$, this follows from Lagrange's Theorem in Group Theory, applied to the multplicative group of $E$, which has order $p^n - 1$. Thus, in fact $g^{p^n} = g$ for all $g\in E$.
Conclusion The polynomial $X^{p^n}- X \in F_p[X]$ has at most $p^n$ roots over $E$, and all of the elements of $E$ are roots by the previous observation, hence $X^{p^n} - X = \prod_{g\in E} (X-g)$.
The previous formula implies immediately that $X^{p^n}- X \in F_p[X]$ is separable, and that $E$ is the splitting field of it! Since $E$ is the splitting field of a separable polynomial over $F_p$, we have that $E/F_p$ is Galois! $(**)$
$(*)$ Indeed, since any element of $E$ can be written uniquely as $a_1 v_1 + \ldots + a_n v_n$ where $\{v_1,\ldots,v_n\}$ is a base for $E$ over $F_p$, and we have $p$ possible choices for each $a_1,\ldots,a_n$, thus $p^n$ choices in total.
$(**)$ This is a general result, whenever you have a field extension $E/F$ and $E$ is the splitting field of a separable polynomial over $F$, then $E/F$ is Galois. (This is problem $8$ from section $(6.3)$ in the pdf)
Comment 1 We just saw that, whenever you have a finite field with $p^n$ elements, we have $g^{p^n} = g$ for all $g$ in the field. In particular, for $n=1$ you have $g^p = g$ for all $g$ in the field (this case is also known as Fermat's Little Theorem). As we saw above, this is a consequence of Lagrange's Theorem.
Comment 2 Also, since splitting fields of a given polynomial are unique up to isomorphism, we get that all finite fields with a given number of elements are isomorphic.