Show that $U_i \rightarrow U_{i+e}, x \mapsto x^p$ is an isomorphism?
Let $K$ be a finite extension of $\mathbb{Q}_p$ with uniformizer $\pi$, prime $\mathfrak p$, and ramification index $e = e(\mathfrak p/p)$. Let $U_i = 1 + \mathfrak p^i$.
If $i > \frac{e}{p-1}$, show that the map $x \mapsto x^p$ is an isomorpism $U_i \rightarrow U_{i+e}$.
Given an element $1 + \pi^{i+e}x \in U_{i+e}$, $x \in \mathcal O_K$, I tried used the following version of Hensel's lemma using the polynomial $f(X) = X^p - 1 - \pi^{i+e}x$, to show the surjectivity of the given map.
Hensel's lemma: if $\alpha_0 \in \mathcal O_K$ satisfies $\nu_{\mathfrak p}(f(\alpha_0)) > \nu_{\mathfrak p}(f'(\alpha_0)^2)$, there is a unique $\alpha \in \mathcal O_K$ such that $f(\alpha) = 0$ and $\nu_{\mathfrak p}(\alpha - \alpha_0) > \nu_{\mathfrak p}(f(\alpha_0)) - \nu_{\mathfrak p}(f'(\alpha_0)^2)$.
Take $\alpha_0 = 1$. Then $\nu_{\mathfrak p}(f(\alpha_0)) = \nu_{\mathfrak p}(-\pi^{i+e}x) \geq i + e$, and $\nu_{\mathfrak p}(f'(\alpha_0)^2) = 2 \nu_{\mathfrak p}(p) = 2e$.
So $f$ should have a $p$th root $\alpha$, with $\alpha - 1 \in \mathfrak p^{i - e}$. But this still doesn't work, because I need a $p$th root in $U_i$. Any hints?