Derivation of $\lim_{n\rightarrow \infty} p^{-n}(x^{p^n}-1)$ in $\Bbb{Q}_p$ Given the definition of $\ln_p(x)$ the $p$-adic logarithm, say by power series, one can show in $\Bbb{Q}_p$ that $\lim_{n\rightarrow \infty} p^{-n}(x^{p^n}-1)=\ln_p(x)$, but
the analogous limit in $\Bbb{R}$, i.e. $\lim_{n\rightarrow \infty} n(x^{1/n}-1)=\ln(x)$, can be derived as a Riemann Sum for $\int_1^x \frac{dt}{t}$ where the interval from $1$ to $x$ is partitioned into unequal subintervals with end points $1, x^{1/n},x^{2/n},...x^{n/n}$, c.f. Burk The Logarithm Function and Riemann Sums in The College Mathematics Journal, Vol 32 No. 5 November 2001 or also this StackExchange post https://math.stackexchange.com/a/3002438/254075 for the same idea of partitioning the interval of integration using a geometric progression.
I was looking for a similar derivation in $\Bbb{Q}_p$ but can't seem to make things work.  Proceeding naively and without any real motivation you can consider the geometric progression $1, x^{p^n},x^{2p^n},...,x^{p^n\cdot p^{n}}=x^{p^{2n}}$ and form sort of a pseudo-Riemann Sum
$$
\begin{align}
(x^{p^n}-1) \cdot \frac{1}{1} &=(x^{p^n}-1) \cdot \frac{1}{1} \\
(x^{2p^{n}}-x^{p^n}) \cdot \frac{1}{x^{p^n}} &=x^{p^n}(x^{p^n}-1) \cdot \frac{1}{x^{p^n}} \\
(x^{3p^{n}}-x^{2p^{n}}) \cdot \frac{1}{x^{2p^{n}}} &=x^{2p^n}(x^{p^n}-1) \cdot \frac{1}{x^{2p^n}}\\
& \vdots
\end{align}
$$
Adding up the terms leads to $p^{n}(x^{p^n}-1)$, since there are $p^{n}$ terms.  But what is needed is $p^{-n}(x^{p^n}-1)$, and what would be the motivation for such a procedure anyway?  Does anyone have thoughts, ideas, hints or references on an integral and a $p$-adic measure that could be used to derive $\lim_{n\rightarrow \infty} p^{-n}(x^{p^n}-1)$ and thereby define $\ln_p(x)$?
 A: We can define the p-adic logarithm through the Volkenborn integral this way,
$$\ln_p(x) = \int_{\mathbb{Z}_p} x^{y+1}-x^y dy = \lim_{n \to \infty} \frac{1}{p^n} \sum_{k=0}^{p^n-1} x^{k+1}-x^k =  \lim_{n \to \infty} \frac{x^{p^n}-1}{p^n}$$
I felt like the question was wanting a little bit more than that, so consider the rest as bonus content to play around with. I figured it would be interesting to try to prove the regular logarithm rules we know and love starting here as our definition, for instance we can clearly see $\ln_p(1)=0$.
$$\ln_p(x) = (x-1)\int_{\mathbb{Z}_p} x^y dy$$
I wasn't able to get $\ln(ab)=\ln(a)+\ln(b)$ however we can get $\ln(x^n)=n \ln(x)$ when $|x-1|_p<1$ using a property of Volkenborn integrals:
$$\int_{\mathbb{Z}_p} f(z)dz = \frac{1}{n}\sum_{i=0}^{n-1} \int_{\mathbb{Z}_p} f(i+nz)dz$$
Now we have,
$$\int_{\mathbb{Z}_p} x^zdz = \frac{1}{n}\sum_{i=0}^{n-1} \int_{\mathbb{Z}_p} x^{i+nz}dz$$
$$\frac{\ln x}{x-1} = \frac{1}{n}\int_{\mathbb{Z}_p} (x^n)^zdz\sum_{i=0}^{n-1} x^i $$
$$\frac{\ln x}{x-1} = \frac{1}{n}\frac{\ln(x^n)}{x^n-1}\frac{x^n-1}{x-1}$$
$$\ln x = \frac{1}{n}\ln(x^n)$$
Furthermore while we're here, since $\ln(1)=0$ and the set of $x$ such that $|x-1|_p<1$ contains the $p$-power roots of unity in $\mathbb{C}_p$, this implies that for $\zeta^{p^n}=1$:
$$\ln \zeta = \frac{1}{p^n} \ln(\zeta^{p^n}) = \frac{1}{p^n}\ln(1) = 0$$
I suppose it might be possible either through a bit more careful integral maneuvering to get the more general $\ln(ab)=\ln(a)+\ln(b)$ or maybe we can say for any $a,b$ that we can find some sequence of the form $\ln(x^{m_i} x^{n_i}) = \ln(x^{m_i})+\ln(x^{n_i})$ with $x^{m_i} \to a$ and $x^{n_i} \to b$. These are just some possible ideas, don't quote me on it if they don't pan out!
You can find more discussion about the Volkenborn integral in many books such as Koblitz's, Robert's, or Schikhof's.
