$p$-adic logarithm, $|\log_p(1 + x)|_p = |x|_p$? Define the $p$-adic logarithm$$\log_p(1 + x) = \sum_{i =1}^\infty (-1)^{i-1}x^i/i.$$How do I see that if $p > 2$ and $|x|_p < 1$, then $|\log_p(1 + x)|_p = |x|_p$?

Here is an outline of a proof.  First recall that the disc of convergence of $\log_p(1+x)$ is $D(0,1) = p\mathbb{Z}_p$.  Let $v_p$ denote the $p$-adic valuation.  Given $x \in p\mathbb{Z}_p$, then $v_p(x) \geq 1$.
($1$) Show that $v_p\left(\frac{x^n}{n}\right) > v_p(x)$ for all $n \geq 2$.  Intuitively, this should hold because $x$ has at least $1$ factor of $p$, so $x^n$ has at least $n$ factors of $p$, while the number of factors of $p$ in $n$ grows at most logarithmically.  More formally, write $n = up^k$ for some $k \in \mathbb{Z}_{\geq 0}$ and unit $u \in \mathbb{Z}_p^\times$.  Then
\begin{align*}
v_p\left(\frac{x^n}{n}\right) = n v_p(x) - v_p(n) = up^k v_p(x) - k \, .
\end{align*}
($2$) Recall that $v_p(a \pm b) \geq \min\{v_p(a),v_p(b)\}$.  Show if $a,b \in \mathbb{Q}_p$ and $v_p(a) \neq v_p(b)$, then $v_p(a \pm b) = \min\{v_p(a),v_p(b)\}$.
$(3)$ Consider a partial sum
$$
s_N = \sum_{n = 1}^N (-1)^{n-1} \frac{x^n}{n} = x - \frac{x^2}{2} + \frac{x^3}{3} - \cdots \pm \frac{x^N}{N} \, .
$$
Since $v_p\left(\frac{x^n}{n}\right) > v_p(x)$ for all $n \geq 2$ by $(1)$, then
$$
v_p(s_N) = \min\left\{v_p(x), v_p\left(\frac{x^2}{2}\right), \ldots, v_p\left(\frac{x^N}{N}\right)\right\} = v_p(x)
$$
by $(2)$, so $|s_N|_p = |x|_p$ for all $N$.
$(4)$ Since the absolute value $|\cdot|_p$ is continuous, then
$$
|x|_p = \lim_{N \to \infty} |s_N|_p = |\lim_{N \to \infty} s_N|_p = |\log_p(1+x)|_p
$$
as desired.
