$f$ is irreducible in $\Bbb F[x]$ 
Let $\Bbb F$ be a field of characteristic $p\gt 0$ and $f(x)=x^{p^n}-c \in\Bbb F[x]$ where $n$ is a positive integer. If $c \notin \{a^p:a\in \Bbb F \}$, show that $f$ is irreducible in $\Bbb F[x]$.

I recently started studying Field theory, so I don't know How to approach this problem? Any help would be appreciated.
 A: Let $\gamma\in \mathbb F^{alg}$ be a root of $f(x)=x^{p^n}-c \in\Bbb F[x]$ in an algebraic closure of $\mathbb F$.
 Thus we have $f(x)=x^{p^n}-c=(x-\gamma)^{p^n}$.
On the other hand let $m(x)\in \Bbb F[x]$ be the minimal polynomial of $\gamma$ over  $\Bbb F$.  
If $n(x)$ is an irreducible monic factor of $f(x)$ over $\Bbb F$, its only root in $\mathbb F^{alg}$ is $\gamma$, so that $n(x)$ must also be the minimal polynomial of $\gamma$ in $\Bbb F[x]$. Hence  $n(x)=m(x)$.  
This means that the decomposition of $f(x)$ into irreducibles in  $\mathbb  F[x]$ is $f(x)=m(x)^{p^e}$ for some integer $e\geq 0$ [the exponent is a power of $p$ because it divides $p^n=deg (f(x))$].
More explicitly we have  $f(x)=x^{p^n}-c =m(x)^{p^e}=(x^s+...+m_0)^{p^e}$.
But then $c=-m_0^{p^e}=(-m_0)^{p^e}$ and since $c$ is supposed not to be  a $p$th-power in $\Bbb F$, this forces $e=0$ and so $f(x)=m(x)^{p^e}=m(x)$ is irreducible in  $\Bbb F[x]$
A: Go up to a splitting field $E$ so that in there $f(x)$ splits completely. Now a root of $f(x)$ is $\sqrt[p^n]{c}$ and so we can write $f(x)$ over $E$ as 
$$f(x) = x^{p^n} - (\sqrt[p^n]{c})^{p^n} = (x - \sqrt[p^n]{c})^{p^n}$$
by the schoolboy binomial theorem. Now recall $f(x) \in \Bbb{F}[x]$ so that each coefficient of $f(X)$ is an element of $\Bbb{F}$. Suppose that $f(x)$ is reducible. From the factorisation above, we see that any divisor of $f(x)$ must be of the form 
$$(x - \sqrt[p^n]{c})^{r}$$
where $0 \leq r \leq p^n$. Now expand this out by the ordinary binomial theorem and look at the coefficient of the second highest power of the indeterminate $x$. Can you deduce the irreducibility of $f(x)$ from here? Hint: show that $r = 0$ or $r = p^n$.
