Artin Schreier equation Let $K$ be the field obtained by adjoining to $\mathbb{Q}_p$ a root $\alpha$ of the polynomial $f(x)=x^p-x-\dfrac{1}{p}$. I should prove that $K \supset \mathbb{Q}_p$ is a Galois extension of degree $p$. I managed to prove that $K$ has degree $p$:taking $\dfrac{1}{\alpha}$ we see that is a root of $x^p-px^{p-1}-p$ which is an Eisenstein polynomial.
I do not know how to show that we have all roots of $f$.

It's just Hensel's lemma.
First note that the $p$-adic valuation of $\alpha$ is $-1/p$: from the identity $\alpha^p-\alpha=1/p$, it follows that $v_p(\alpha) <0$, and then by triangle inequality $v_p(\alpha^p)=v_p(\alpha^p-\alpha)=-1$.
Putting $K = \mathbb{Q}_p(\alpha)$ and $y = x - \alpha$, we want to show that the polynomial $g(y) = (y + \alpha)^p - (y + \alpha) - \frac{1}{p} \in K[y]$ has $p$ roots in $K$.
Now write $g(y) = y^p + c_{p - 1}y^{p - 1} + \cdots + c_2y^2 + (c_1 - 1)y$, where every $c_i$ is in the maximal ideal of $\mathcal{O}_K$. In fact, every $c_i$ is of the form $p\cdot d_i \cdot \alpha^{p-i}$, where $d_i$ is an integer. We thus have $v_p(p\cdot \alpha^{p-i})= 1-(p-i)/p=i/p>0$.
In the residue field, we have $\overline{g}(\overline{y}) = \overline{y}^p - \overline{y}$, which has $p$ different roots, namely the elements of $\mathbb{F}_p$.
Thus by Hensel's lemma, for every element of $\mathbb{F}_p$ there is a unique root of $g$ in $K$ lifting that element.
