A question about the $p$-adic product formula for $\log(1+X)$ and $p$-adic geometry Fix a prime number $p$. There is a well known product formula for $\log(1+X)$ as a power series in $\mathbb{Q}_p[[X]]$ (by Serre, I think). Namely letting $\Phi_n(X)=(1+X)^{p^n}-1$ and $Q_n(X)=\Phi_{n+1}(X)/p\Phi_n(X)$, then we have$$\log(1+X)=X\prod_{n\geq1}Q_n(X).$$
On the other hand, it is well known that the ring $\mathbb{Q}_p[[X]]$ is noetherian. Suppose the elements $\log(1+X)$, $\log(1+X)/X$, $\log(1+X)/XQ_1(X)$, etc. all exist (in the sense that the products converge). Then the ideals defined by these elements  must become larger and larger, and so they must stabilize. On the other hand, it seems like the number of zeros in $\mathbb{C}_p$ of each such element becomes strictly smaller at each step (as we are dividing by the minimal polynomial of the primitive $p^n$-power roots of unity shifted by 1), so something must be wrong here, unless these products don't all converge. So it seems like from some point onwards, if we remove enough factors, this product will not converge to an element of $\mathbb{Q}_p[[X]]$.
My question is: what exactly is happening here? Is it true that the product stops converging after we remove finitely many factors, and when does it happen? Or am I missing something simple?
Also, I would like to understand what is happening from a geometric perspective. Namely, it seems that $\mathbb{Q}_p[[X]]$ is the ring of functions on the open rigid analytic unit disc over $\mathbb{Q}_p$. Then the zeros of $\log(1+X)$ are the $p$-power roots of unity, shifted by 1. Thus there are infinitely many zeros in the disc, and they diverge to its boundary. But it seems what happens above must mean that we can't take an arbitrary tail of such points and expect it to define the zero set of some ideal (contrary to the situation in algebraic geometry). So, what is happening? When does a closed subset correspond to an ideal of the ring $\mathbb{Q}_p[[X]]$?
All of this can be contrasted with the case of ideals in 1-dimensional affinoid algebras, which correspond to vanishing loci in closed discs. In that case functions can only have finitely many zeros, by the Weierstrass preparation theorem, so the situation is akin to what happens in algberaic geometry.
Thanks!
 A: Not sure if this really answers your question, but here's at least a derivation of the formula (assuming the product should be for all $n\ge0$ rather than $n\ge1$).
First, using $\Phi_0(X)=X$, note that the partial products
$$
\Psi_n(X)
= X\cdot\prod_{k=0}^{n-1} Q_k(X)
= \Phi_0(X)\cdot\prod_{k=0}^{n-1}\frac{\Phi_{k+1}(X)}{p\Phi_k(X)}
= \frac{\Phi_n(X)}{p^n}
= \frac{(1+X)^{p^n}-1}{p^n}.
$$
So, the claim is that $\Psi_n(X)\rightarrow\log(1+X)$ as $n$ increases.
Let's expand
$$
\Psi_n(X)
= \frac{(1+X)^{p^n}-1}{p^n}
= \sum_{i=1}^{p^n} \frac{\binom{p^n}{i}}{p^n} X^i
= \sum_{i=1}^{p^n} \frac{(p^n-1)\cdots(p^n-i+1)}{1\cdots(i-1)}\cdot\frac{X^i}{i}.
$$
It is tempting to reduce this modulo $p^n$ and expect the terms to become $(-1)^{i-1}X^i/i\mod p^n$, but for $i\ge p$ there will be multipla of $p$ in the denominator. However, if we look at terms of degree at most $k$ modulo $p^m$, the terms will remain constant for sufficiently large $n$: I think $n\ge m+k/(p-1)$ should suffice, where the extra margin is to overcome the number of times $p$ divides the denominator.
A: As you recognize in a comment, the right ring is the subring of $\Bbb Q[[x]]$ consisting of all $f=\sum_0^\infty a_ix^i$ for which $\liminf_iv(a_i)/i=0$. These are precisely the $\Bbb Q$-power series that give you a convergent series when you substitute $x\to\xi$, no matter the $\xi\in\Bbb C_p$ with $v(\xi)>0$. Here, of course, I’m using the additive valuation $v$ on $\Bbb C_p$, and we might as well normalize so that $v(p)=1$. I think that for the purpose of this answer, I’ll call our ring $\Bbb Q_p\{\{x\}\}$.
What I want to do, because that’s more comfortable for me, is move the focus of attention from $1$ to $0$, so that instead of multiplying $\xi$ and $\eta$ elements close to $1$, to get $\xi\eta$, I’ll call them $1+x=\xi$ and $1+y$, and “multiply” them as $(1+x)(1+y)-1=x+y+xy$. This is the formal group of multiplication. I’ll write $M(x,y)=x+y+xy$, and define an endomorphism of $M$ as any $\Bbb C_p$-series, $\psi$, with $p$-integral coefficients, such that $M\bigl(\psi(x),\psi(y)\bigr)=\psi\bigl(M(x,y)\bigr)$. It’s a wonderful fact that all $M$-endomorphisms are in $\Bbb Z_p[[x]]$, and that for every $\gamma\in\Bbb Z_p$, there is an endomorphism $\psi$ with $\psi'(0)=\gamma$. One calls this endomorphism $[\gamma]_M$. You see immediately that for any positive integer $n$, $[n]_M(x)=(1+x)^n-1$, in particular for $n=p^m$. These are the polynomials that appear in @EinarRødland’s answer.
I’ll use the same shift of coordinates to call $\text{Log}(x)=\sum_{n=1}^\infty(-1)^{n+1}x^n/n$. This counts as a homomorphism from $M$ to the additive formal group $A(x,y)=x+y$, that is, $\text{Log}\bigl(M(x,y)\bigr)=\text{Log}(x)+\text{Log}(y)$.
You do see that for $z\in\Bbb Z_p$, $(1+x)^z-1$ makes sense all right, ’cause for each $i\ge1$, the binomial coefficient $\binom zi$ is in $\Bbb Z_p$; here’s one way of showing this. Since $z\in\Bbb Z_p$, there’s a sequence $\{n_j\}$ of positive integers $n_j$ with $z$ as limit. The associated binomial functions
$\binom xi=x(x-1)(x-2)\cdots(x-i+1)/i!$ are polynomials and thus continuous as
functions $\Bbb Q_p\to\Bbb Q_p$, and take $\Bbb Z$-values when $x$ is evaluated to any positive integer. Thus $\binom zi\in\Bbb Z_p$.
Perhaps without noticing it, we used the topology of coefficientwise convergence  in $\Bbb Q\{\{x\}\}$ in the preceding argument. (If you look closely enough, this is also the topology of uniform convergence on proper subdiscs of the unit disk in $\Bbb C_p$.) We can use the same notion of convergence to show that
$$
\text{Log}(x)=\lim_m\frac{[p^m]_M(x)}{p^m}\,.
$$
I  was hoping to show this by looking at the binomial coefficients, but easiest seems to be to differentiate each $[p^m](x)/p^m$ to get $(1+x)^{p^m-1}$, which have limit $(1+x)^{-1}$, $p$-adically, coefficientwise.
You look at $[p^m](x)/p^m$, and realize that it’s a finite product
$$
\frac{[p^m](x)}{p^m} = x\prod_{k=1}^m\frac{[p^k](x)}{p[p^{k-1}](x)}\,,
$$
in which the factors are polynomials that have limit $1$, $p$-adically, coefficientwise. Of course
in the first factor, $[p^0](x)=x$, the identity endomorphism. This is exactly what @EinarRødland has said.
I’ve left out a lot of details, I know. But it all works out fine, and gives you,
I think, a clear idea of what the roots of the logarithm Log are. This should answer most of your questions. But I should point out that the roots of the Log, i.e. the numbers $\zeta_{p^n}-1$ do not “converge to the boundary”, although their valuations $v(\zeta-1)$ do converge to zero. It’s an entirely different thing. The roots of unity in $\Bbb C_p$ are a discrete group, in strong contrast to the complex situation.
A: For $x \in p\Bbb{Z}_p$
$$\log(1+x) = \sum_{k=1}^\infty \frac{(-1)^{k+1}}{k}x^k, \qquad \exp(p^nx) =\sum_{k=0}^\infty \frac{p^{nk}x^k}{k!}= \exp'(p^nx)$$ $$ (1+x)^{p^n} = \exp(p^n \log(1+x)), \qquad g_x(y) = \exp(y \log(1+x))$$ $$ \log(1+x) =g_x'(0)=\lim_{n \to \infty} \frac{\exp(p^n \log(1+x))-1}{p^n}=\lim_{n \to \infty} \frac{(1+x)^{p^n}-1}{p^n}$$ $$= \lim_{n \to \infty} x\prod_{m=1}^n \frac{(1+x)^{p^m}-1}{p((1+x)^{p^{m-1}}-1)}$$
With $\zeta_{p^m}\in \overline{\Bbb{Q}}_p$ a primitive $p^m$-th root of unity, the roots of $(1+x)^{p^m}-1$ are the $\zeta_{p^m}^a-1, a \in 0 \ldots p^m-1$ so that  $$\frac{(1+x)^{p^m}-1}{p((1+x)^{p^{m-1}}-1)}= \prod_{a=1,\ p\ \nmid\ a}^{p^m} (1-\frac{x}{\zeta_{p^m}^a-1})\in \Bbb{Q}_p[x], \qquad  v_p(\zeta_{p^m}^a-1) = \frac1{(p-1)p^{m-1}}$$
which means $1-\frac{x}{\zeta_{p^m}^a-1}$ looks like $1-xp^{-\epsilon}$ and it is only the product of all its conjugates that $\to 1$.
Because $\frac{(1+x)^{p^n} -1}{p^n } \in x+x^2 \Bbb{Q}_p[[x])$ the ideal it generates is $(x)$. 
