Prove that a polynomial is irreducible or the field contains a $p$th root An exercise from Lang:

Let $k$ be a field. Let $n$ be a nonnegative integer.
Consider the polynomial $x^{p^n}-a \in k[x]$.
Assume that $\text{char}(k) = p > 0$.
Prove that either the polynomial $x^{p^n}-a$ is irreducible over the field $k$ or there exists a $b\in k$ such that $a = b^p$.

How to prove this statement? I am a little bit confused with this statement.
 A: One side:
If $a=b^{p}$ for $b\in k$ then: $$x^{p^{n}}-a=x^{p^{n}}-b^{p}=\left(x^{p^{n-1}}-b\right)^{p}$$
showing that the polynomial is not irreducible.

Let $k^{\mathbf{a}}$ denote an algebraically closed fied that contains
$k$ and is algebraic over $k$.
Let it be that $\alpha\in k^{\mathbf{a}}$ is a root of $x^{p^{n}}-a$. 
Then $\alpha^{p^{n}}-a=0$ so that $x^{p^{n}}-a=x^{p^{n}}-\alpha^{p^{n}}=\left(x-\alpha\right)^{p^{n}}$. 
This reveals that polynomial $x^{p^{n}}-a$ has only one root in $k^{\mathbf{a}}$. 

The other side:
For convenience define $f\left(x\right):=x^{p^{n}}-a$ and let $g\left(x\right)\in k\left[x\right]$
denote the minimal polynomial of $\alpha$. 
Then $f\left(x\right)=g\left(x\right)^{m}h\left(x\right)$
for some positive integer $m$ and some $h\left(x\right)\in k\left[x\right]$.
Taking $m$ maximal $g\left(x\right)$ will not divide $h\left(x\right)$
so that $\alpha$ is not a root of $h\left(x\right)$. But $\alpha$
is the only candidate for being a root of $h\left(x\right)$ so we
conclude that $h\left(x\right)$ has no roots at all. This allows
the conclusion that: $$f\left(x\right)=g\left(x\right)^{m}$$ If $d$
denotes the degree of $g$ then $md=p^{n}$ showing that $m$ and
$d$ are both powers of $p$. So nonnegative integers $u,v$ exist
with $u+v=n$, $d=p^{u}$ and $m=p^{v}$. Then: $$x^{p^{u}}-\alpha^{p^{u}}=\left(x-\alpha\right)^{p^{u}}=g\left(x\right)\in k\left[X\right]$$
Proved is now that $\alpha^{p^{u}}\in k$. 
If $f$ is not irreducible then $p^{v}=m>1$ so that $v\geq1$. 
Defining $b:=\alpha^{p^{n-1}}=\left(\alpha^{p^{u}}\right)^{p^{v-1}}\in k$
we find:
$$a=\alpha^{p^{n}}=b^{p}$$
A: Rename the field $k$ as $K$. Rename the indeterminate $x$ as $X$. Thus we must prove that either the polynomial $X^{p^n}-a$ is irreducible over $K$ or $a$ is a $b$-th power.
In an extension field of $K$, if $b$ is a root of $X^{p^n}-a$, then $(X-b)^{p^n}=X^{p^n}-b^{p^n}
=X^{p^n}-a$. Therefore, any proper monic factor of $X^{p^n}-a$ is
$(X-b)^k$ where $0<k<p^n$. If $(X-b)^k$ has coefficients in $K$, then $b^k\in K$.
But as $b^{p^n}=a\in K$, then $b^g\in K$, where $g=\gcd(k,p^n)$. Then $g=p^m$
where $0\le m\le n-1$, and this implies $b^{p^{k-1}}\in K$, and that's a $p$-th
root of $a$.
