I know that the $p$-adic integral defined by $$ \log_p(1 + x) = \sum_{n=1}^{\infty} \frac{(-1)^{n+1} x^n}{n} $$ is well defined for $x \in \mathbb{C}_p$, the completion of the algebraic closure of $\mathbb{Q}_p$, with $|x|_p < 1$.
I am wondering how does one see that $\log_p(1 + x) \in \mathbb{Z}_p$ given $x \in p \mathbb{Z}_p$?
On
You first need the following fact on series in the $p$-adics:
The series $\sum_{n=1}^\infty a_n$ converges if and only if $|a_n|_p \to 0$ as $n \to \infty$.
The sufficient part is of course incorrect if we take the normal norm on $\mathbb R$, but this is true for non-archimedean norms, like the $p$-adic norm.
Let $a_n = \frac{(-1)^{n+1}}{n}x^n$. Then $|a_n|_p = |n|_p^{-1}|x|_p^n \leq n |x|_p^n$, where the equality is attained only when $n = p^l$ for some $l \in \mathbb Z_{\geq0}$. So $|a_n|_p$ has limit $0$ whenever $|x|_p < 1$.
This is easier than it looks:
Using additive valuation, for which $v_pp=1$, so that $\vert z\vert_p=p^{-v_pz}$, we have $v_p(x^n)\ge n$, while $v_p(n)\le\log_p(n)$. Thus $v_p(x^n/n)\ge1$, so the sum is in $p\Bbb Z_p$.