$X^{p^k} - a ∈ K[X]$ irreducible? Let $K$ be a field of characteristic char$(K) = p > 0$, and let $a ∈ K$ be an element with the following property:
$$(\forall \beta ∈ K)(\beta^p ≠ a).
$$
Let $k ∈ ℕ$ be arbitrarily given. Is it then true that $X^{p^k} - a$ is irreducible in $K[X]$?
We know that in the extension $K ⊆ K[X]/(X^{p^k} - a) \cong K(\sqrt[p^k]{a})$, that if $X^{p^k} - a$ were irreducible, then it would be the minimal polynomial of $\sqrt[p^k]{a}$ over $K$. Is this something we can use? 
 A: In the following, I use 'inseparable' to mean a polynomial has a repeated root.
An important fact is that, if $F$ is a field of characteristic $p>0$, then a polynomial $f(x)\in F[X]$ is irreducible and inseparable iff it equals $g(x^p)$ for some $g(x)\in F[X]$.
Now say that some $a\in K, \mathrm{char}(K)>0$ has no $p$th root in $K$. Suppose by way of contradiction then that some $x^{p^k}-a$ is reducible over $K$. $x^{p^0}-a=x-a$ is clearly irreducible, so $k>0$. Let $j$ be the smallest nonnegative value of $k$ for which $x^{p^k}-a$ is reducible. Clearly we must have $j>0$. Now write $x^{p^j}-a$ as a product of its monic irreducible factors in $K[X]$, $x^{p^j}-a=\prod_{i=1}^n f_i(x)$. 
If $\sqrt[p^j]{a}$ is a $p^j$th root of $a$, then $x^{p^j}-a$ factors as $(x-\sqrt[p^j]{a})^{p^j}$ (in the extension $F(\sqrt[p^j]{a})$) so each $f_i(x)$ factors as some $(x-\sqrt[p^j]{a})^{c_i}$. If some $c_k=1$, then $f_k(x)=x-\sqrt[p^j]{a}$ which means $\sqrt[p^j]{a}\in K$, but this is impossible as $K$ lacks even a $p$th root of $a$. So all $c_i$ are $>1$. This means each $f_i(x)$ is inseparable. As each $f_i(x)$ is, by assumption, irreducible, it follows that there exist a sequence of polynomials, $g_i(x)\in K[X]$, such that $f_i(x)=g_i(x^p)$.  
So $x^{p^j}-a=\prod_{i=1}^n g_i(x^p)$. Now, substituting $y=x^p$, we get $y^{p^{j-1}}-a=\prod_{i=1}^n g_i(y)$ (this is valid as $j>0$). By the way all the $g_i$s were chosen, it is clear that they all have degree $>0$. Combining this with the fact that $n>1$, we have that $x^{p^{j-1}}-a$ is also reducible, a contradiction.  
So all polynomials of the form $x^{p^k}-a$ are irreducible over $K[X]$.  
(please comment or edit any corrections or suggestions)
