Compute Galois group of $\mathbb F_q/\mathbb F_p$ I have to compute Galois group of $\mathbb F_q/\mathbb F_p$ where $q=p^n$. I know already that $\mathbb F_q/\mathbb F_p$ is galois, so I don't need to prove it. Moreover, I know that \begin{align*}\sigma :\mathbb F_q&\longrightarrow \mathbb F_q\\ x&\longmapsto x^p\end{align*}
is an automorphism that fix elements of $\mathbb F_p$ and thus $\sigma \in Gal(\mathbb F_q/\mathbb F_p)$.
Now, I know that if $k<p^n$ then $\sigma ^k(x)=(x^{p})^k=x^{pk}\neq 1$ and thus $\sigma $ is of order $p^n$. Now if I can prove that $|Gal(\mathbb F_q/\mathbb F_p)|=p^n$, then $Gal(\mathbb F_q/\mathbb F_p)=\left<\sigma \right>$ and it would be finish, but I can't prove it since even $X^q-X$ split over $\mathbb F_q$, theis polynomial is not irreducible, and thus, I can't conclude. So how can I find an irreducible polynomial of degree $q$ that split over $\mathbb F_q$ ?
 A: There is a bit of confusion in your attempt but some ideas are alright. Let me try to set this straight. 
The degree of the extension is $n$. Thus the Galois group has $n$ elements.
You know one, the Frobenius automorphism $\sigma$. Now, for $k < n$ (not $p^n$) one has that $\sigma^{k}(x) = x^{p^k}$ (not $x^{pk}$, note that $\sigma(\sigma(x))= \sigma(x^p)=(x^p)^p = x^{p^2}$ and so on), and 
then, since $x^{p^k} \neq x$ (not $1$) one concludes $\sigma^k$ is not the identity map (that is it is not the "one-element" of the Galois group).
Consequently the order of $\sigma$ is (at least) $n$, and the $n$ automorphisms $\sigma^{0}, \sigma^1, \dots, \sigma^{n-1}$ are all distinct.  Thus this must be the full Galois group. That is, the Galois group is cyclic of order $n$; more precisely, it is generated by the Frobenius automorphism.  
A: The exponent $k$ in $\sigma^k$ accounts for the number of compositions of the function $\sigma$ with oneself :
$$\sigma^k := \underbrace{\sigma \circ \dots \circ \sigma}_{k \ \textrm{times}} $$
Hence, the function $\sigma^k$ corresponds to the polynomial $X^{p^k}$, ie $\sigma^k(x) = x^{p^k}$.
The polynomial you're looking for is simply $P := X^q - X$.
Proof: $P$ is a polynomial of degree $q$ whose $q$ roots are precisely the elements of the fixed field of $\sigma^n$, namely $\mathbb{F}_q$. One hence gets the factorization :
$$P = \prod_{a \in \mathbb{F}_q} (X-a) $$
