I know that the polynomial $f(x)=x^p-x-\frac1p\in\mathbb Q_p[x]$ is irreducible. So, let $\alpha$ be a root of $f(x)$, and $K=\mathbb Q_p(\alpha)$. Let $O_K$ be the valuation ring of $K$ with respect to the unique extension of the p-adic valuation to $K$. Let $\mathfrak p_K$ be the unique maximal ideal of $O_K$. We know that $O_K$ is also the integral closure of $\mathbb Z_p$ in $K$.
I am trying to show that there exists $\beta_i \in O_K$ for $i =0,...,p-1$ such that $\alpha + \beta_i$ is a root of $f(x)$ and $\beta_i \equiv i$ (mod $\mathfrak p_k$).
My guess is the following:
I know from Hensel's lemma that the polynomial $x^p - x$ splits in $\mathbb Z_p$ with roots $\beta_i \in \mathbb Z_p$ for $i =0,...,p-1$ such that $\beta_i \equiv i$ (mod $p\mathbb Z_p$). So clearly, $\beta_i \equiv i$ (mod $\mathfrak p_k$). What remains to show is that $\alpha + \beta_i$ are the roots that I seek.
I think I am on the right track, but I don't know how to show that $\alpha + \beta_i$ are roots of $f(x)$. Any help or hint would be appreciated.
Of course, my guess could be totally wrong. In that case I have no idea how to come up with the desired $\beta_i 's$.