Definition of $p$-adic $(1+x)^\alpha$ via binomial series and log/exp It seems like there are two ways to make sense out of powers in the $p$-adic context, so I wonder if there's any difference. Here is some background (I hope the experts will excuse me for writing out certain obvious things, and I also hope there are no mistakes).


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*First of all, we have the following formal identity in the ring of power series $\mathbb{Q} [[X,Y]]$:
$$\tag{0} \sum_{n\ge 0} {Y\choose n}\,X^n = \exp (Y\cdot \log (1+X)),$$
where the "binomial coefficients" are defined as polynomials
$${Y \choose n} = \frac{Y\,(Y-1)\cdots (Y-n+1)}{n!} \in \mathbb{Q} [Y].$$

*Let $\alpha\in\mathbb{Z}_p$. Then also ${\alpha \choose n} \in \mathbb{Z}_p$ for all $n = 0,1,2,3,\ldots$, and we see that for $|x|_p < 1$ the binomial series
$$\tag{1} (1+x)^\alpha = \sum_{n\ge 0} {\alpha\choose n} \, x^n$$
converges.

*Under the same assumptions, we may consider the $p$-adic series
$$\tag{2} \exp (\alpha\cdot \log (1+x)).$$
However, even though $\log (1+x)$ again converges for $|x|_p < 1$, the $p$-adic exponential series converges only for $|x|_p < p^{-1/(p-1)}$, and $\log$ and $\exp$ are inverse to each other only as functions $1 + B (0,p^{-1/(p-1)}) \leftrightarrow B (0,p^{-1/(p-1)})$, so I suppose we have to restrict our attention to $|x|_p < p^{-1/(p-1)}$ to make sure that everything works.
Here's a couple of questions:


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*What is the easiest way to show that (1) and (2) coincide for $|x|_p < p^{-1/(p-1)}$? My suggestion is calculate the Taylor coefficients, which works both formally and non-formally. For $f (X) = \exp (Y\cdot \log (1+X))$ we have
$$f^{(n)} (X) = \exp (Y\cdot \log (1+X))\cdot\frac{Y\,(Y-1)\cdots (Y-n+1)}{(1+X)^n},$$
so we see that the coefficients $f^{(n)} (0)/n!$ are exactly what we need... I guess it is easy enough?

*Which of the two definitions is better: (1) or (2)? Well, if I understand well (?), (1) has a bigger domain of convergence. Is that correct?
Note that if we are interested in $\mathbb{Q}_2$ and $\alpha = 1/2$ (the typical application to finding square roots), then it seems like both approaches give the same result. In the denominators of ${1/2 \choose n}$ we have powers of $2$; namely,
$$\left|{1/2 \choose n}\right|_2 = |1/2|^n_2\cdot |n!|_2 = 2^{2n-s_2 (n)},$$
meaning that the series converges for $x\in 8\mathbb{Z}_2$. Similarly, if we consider
$$\exp (1/2\cdot \log (1+x))$$
then it also converges precisely for $x\in 8\mathbb{Z}_2$ (the logarithm converges on $4\mathbb{Z}_2$, but then there is the $1/2$ factor which leads to some troubles with convergency of the exponential, unless we take $x \in 8\mathbb{Z}_2$)... Well, it's not like we should expect something different, knowing what the squares in $\mathbb{Z}_2$ are.
Thank you for your answers and comments.
 A: We’re dealing with a principal unit $z=1+x$ where $|x|<1$, or equivalently $v_p(x)>0$; and with a $p$-adic integer $\alpha$. You ask about two methods of calculating $z^\alpha$, but you don’t say anything about the definition of this construct.
For my money, $z^\alpha$ is to be defined this way: Let $\{n_i\}$ be a sequence of positive integers with $p$-adic limit $\alpha$. Then $z^\alpha$ is defined to be $\lim_iz^{n_i}$. Of course you have to verify that the sequence $\{z^{n_i}\}$ is $p$-adically convergent, and that the result does not depend on the choice of the sequence $\{n_i\}$. From a computational standpoint, you might find that using this definition directly is a faster way of calculating $z^\alpha$ than the Binomial. For your sequence of positive integers, just use what you get by cutting off the $p$-ary expansion of $\alpha$.
I guess that your task is to show that the Binomial series approximation to $z^\alpha$ agrees with the definition for all $z$ in the open unit disk. As you recognize, any method involving the exponential series will be of much lesser applicability.
One other remark: if you’re only calculating $z^{m/n}$, surely the fastest method of computation is to use Newton-Raphson on $X^n=z^m$.
