Local analycity of p-adic power. Let $u \in \mathbb C_p$, $|u|<1$. It is easy to see that the function $n \mapsto (1+u)^n$, $\mathbb N \rightarrow \mathbb C_p$ uniquely extends to a continuous function $f: \mathbb Z_p \rightarrow \mathbb C_p$. One customarily uses the notation $(1+u)^z$ for $f(z)$.
I am looking for an elegant proof, or a reference to any proof, of the following well-known fact:

The function $z \mapsto (1+u)^z$ is locally analytic.

This means that there exists an $r>0$ and coefficients $a_n \in \mathbb C_p$ such that $|a_n|r^n$ goes to  $0$ when $n$ goes to infinity and $(1+u)^z = \sum_{n=0}^\infty a_n z^n$ for all $z \in \mathbb Z_p$, $|z| \leq r$. This should be easy, but the proof I came up with seems way too computationally complicated, especially for the teaching purpose I have.
 A: It’s almost essential to use, instead of absolute value, additive valuation $v=v_p$, normalized so that $v(p)=1$. You’re asking about $(1+u)^z$, for fixed $u$ with $v(u)>0$, as a $\Bbb C_p$-series in $z$.
Now, as a series in two variables, $(1+u)^z\in\Bbb Q_p[z][[u]]$, but we want to look at it as an element of $\Bbb Q_p[[u]][[z]]$. You’ve written this as $\sum_na_nz^n$, suppressing the dependence of the $a_n$’s on $u$.
At least one of the commenters pointed out that it would be nice to know $\frac{\partial^n}{\partial z^n}(1+u)^z$; but we all know that $\frac\partial{\partial z}(1+u)^z=(1+u)^z\log(1+u)$, formally, in $\Bbb Q[[u,z]]$. Therefore,
$$
\frac{\partial^n}{\partial z^n}(1+u)^z=\bigl(\log(1+u)\bigr)^n(1+u)^z\,.
$$
Now, still in $\Bbb Q_p[[u]][[z]]$, we have
$$
(1+u)^z=\sum_n\frac{\bigl(\log(1+u)\bigr)^n}{n!}z^n=\sum_n\frac{\bigl(z\log(1+u)\bigr)^n}{n!}\,,
$$
which all of us knew already, but we didn’t believe it.
Specializing $u$ to a constant, the criterion for convergence of this $z$-series is that $v\bigl(z\log(1+u)\bigr)>1/(p-1)$. You may need to reassure yourself that $v\bigl(\log(1+u)\bigr)$ does not get out of hand. But the dependence of this valuation on $v(u)$ is well understood: if $v(u)\ge1/(p-1)$, then $v(\log(1+u))\ge v(u)$, and if the first inequality is strong, then the inequality on the valuation becomes an equality. Similarly,
$$
\text{if}\quad\frac1{p^{n+1}(p-1)}\le v(u)\le\frac1{p^n(p-1)}\quad
\text{then}\quad v\bigl(\log(1+u)\bigr)\ge p^nv(u)-n\,,
$$
and again, if the conditions are both strongly satisfied, the conclusion becomes an equality.
I think that should do it.
