How can we show that for $x \in \mathbb{Z}_p$, $\log_p(1+x)$ converges in $\mathbb{Z}_p$ when $|x|_p < 1$?
To clarify, $\log_p(1+x)$ is the power series: $$\sum_{n=1}^\infty \frac{(-1)^{n+1}x^n}{n}$$
How can we show that $\exp_p(x)$ converges for: $|x|_p < p^\frac{-1}{p-1}$? How can we approximate the largest power of $p$ dividing $n!$?
$$\exp_p(x) = \sum_{n=0}^\infty \frac{x^n}{n!}$$
I've shown that if $\lim_{n \rightarrow \infty} |a_n|_p = 0$, then $\sum_{n=1}^\infty a_i$ converges. How do I show the following: $$\lim_{n\rightarrow\infty} \left|\frac{(-1)^{n+1}x^n}{n}\right|_p = 0$$ $$\lim_{n \rightarrow\infty} \left|\frac{x^n}{n!}\right|_p = 0$$ I know that: $$\left|\frac{1}{n!}\right|_p = p^\frac{n-S_n}{p-1}$$