Show $x^5+x^4+x^2+x+1$ is irreducible in $\mathbb{Q}(\sqrt[5]{2})$

Question

I have proved that $f(x)=x^5+x^4+x^2+x+1$ is irreducible in $\mathbb{Z}_2$, so it's irreducible in $\mathbb{Z}$, hence in $\mathbb{Q}$.

I know $f(x)=x^5+x^4+x^2+x+1$ is irreducible in $\mathbb{Q}(\sqrt[5]{2})$ $\iff$ $x^5-2$ is irreducible in $\mathbb{Q}(\alpha), f(\alpha)=0 $, but I cannot make progress anymore.

Answer 1

$2$ is totally ramified in $\Bbb Q(\sqrt[5]{2})$, because $\sqrt[5]{2}^5=2$. It follows that $\mathcal O_{\Bbb Q(\sqrt[5]{2})}/(\sqrt[5]{2})=\Bbb F_2 $. Thus by the reduction criterion, irreducibility over $\Bbb F_2 $ implies irreducibility over $\mathcal O_{\Bbb Q(\sqrt[5]{2})}$ and this implies irreducibility over $\Bbb Q(\sqrt[5]{2})$.