Let $K$ be a field with $\operatorname{char}K =p$, where $p$ is a prime number and $a \ne 0,1 \in K $. Assume that $P(x) = x^p-a$ is irreducible in $K[x]$ and let $\alpha$ a root of $P(x)$, show that $x^p - \alpha$ is irreducible over $K(\alpha)$.
I have tried to solve that, but I have some questions:
IF $\operatorname{char}K =p$, then $\operatorname{char}K(\alpha) = p??$
Suppose that $\operatorname{char}K(\alpha) = p$, we have $x^p- \alpha = x^p - (\sqrt[p]{\alpha})^p = (x - (\sqrt[p]{\alpha}))^p$. So the only root is $\sqrt[p]{\alpha}$, but why $\sqrt[p]{\alpha} \notin K(\alpha)$??
Lemma: Let $q=p^k$ and $c\in K$ such that there is no $d\in K$ with $d^p=c$. Then $f(x)=x^q-c$ is irreducible over $K$.
Proof: Let $b$ be a root of $f$ in some extension field. Assume that $f$ factors over $K$, let $g\in K[x]$ be a non-trivial factor of $f$. Since $f(x)=(x-b)^q$ by unique factorization $g$ is of the form $(x-b)^l$ with some $1\leq l<q$. In particular $b^l\in K$. If $l$ was coprime to $p$, then $b=b^{mp+nl}\in K$ (for suitable $n,m\in\Bbb Z$), so $(b^{p^{k-1}})^p=c$ which contradicts your assumption. Otherwise $l=p^n$ for some $n<k$. Then we also have $c=((b^l)^{p^{k-n-1}})^p$, again a contradiction. Hence $f$ is irreducible.
Applying the lemma with $k=1$ we see that it is sufficient to prove $\beta:=\sqrt[p]{\alpha}\notin K(\alpha)$ (as OP already noted). Assume that $\beta\in K(\alpha)$. By assumption the polynomial $x^p-a$ is irreducible over $K$, in particular $a$ has no $p$-th root in $K$ (i.e. $\alpha\notin K$). By the lemma with $k=2$ this implies that $x^{p^2}-a$ is irreducible over $K$. In particular if $z$ is any root of that polynomial in some extension field then $[K(z):K]=p^2$. But $\beta$ is a root of that polynomial and $[K(\beta):K]\leq [K(\alpha):K]=p$ which is a contradiction. Therefore $\beta\notin K(\alpha)$ and hence $x^p-\alpha$ is irreducible over $K(\alpha)$.