How to make sense of the binomial coefficient over $p$-adic integers? I recently asked this question: Show that $y^2=x^3+1$ has infinitely many solutions over $\mathbb Z_p$. and now I'm trying to make sense of the first answer that was posted. It said that I should show $\sum_{n\geq 0}{{1/2}\choose{n}} x^{3n}$ converges, when $p|x$.  
Now because the metric on $\mathbb Z_p$ is non-Archimedian, it is sufficient to show that ${1/2\choose n}x^{3n}\to0$, as $n\to \infty$. This is equivalent to saying that its norm should tend to $0$. So I wrote: 
$$\left\vert{1/2\choose n}x^{3n}\right\vert=\left\vert{1/2\choose n}\right\vert\left\vert x^{3n}\right\vert=p^{-\nu{1/2\choose n}}\cdot p^{-\nu(x^{3n})}.$$ 
Now I get that $p^{-\nu(x^{3n})}\to 0$ if and only if $p|x$. I however don't know what to make of $1/2\choose n$.   
I know that we have this definition: ${k\choose n}=\frac{k(k-1)\cdots (k-n+1)}{n!}$ and that $\frac12$ makes sense in $\mathbb Z_p$, because $p\not\mid2$ (at least for odd $p$). However I don't see why $1/2\choose n$ is well defined, since division is not always possible in $\mathbb Z_p$.  
Also, how can I interpret $1/2\choose n$ in a way, such that I can comprehend $\nu{1/2\choose n}$?
 A: Here’s another approach, which I find helpful for showing that for $p\ne2$, $\binom{1/2}n\in\Bbb Z_p$. We have the binomial functions
$$
\binom tn=\frac{t(t-1)(t-2)\cdots(t-n+1)}{n!}\in\Bbb Q[t]\,.
$$
No question, then, that $\binom{1/2}n$ is a rational, and therefore a $p$-adic number no matter what $n$ is. Let’s see why it’s a $p$-adic integer when $p\ne2$.
For such primes, $1/2$ is a $p$-adic integer, and so is the limit of positive integers. Let $\{m_i\}$ then be a sequence of positive integers with $1/2$ as limit, and look at the $\Bbb Z$-integers $\binom{m_i}n$, which by continuity of the function $\binom tn$, have limit $\binom{1/2}n$. So, since $\binom{1/2}n$ is a limit of $\Bbb Z$-integers, it too is a $p$-adic integer.
Notice that I used no special property of the number $1/2$ beyond its membership in $\Bbb Z_p$. The same proof shows that for a $p$-adic integer $a$, the expression $(1+z)^a$ makes sense for all $p$-adic quantities $z$ for which $|z|<1$.
A: ${1/2 \choose n}$ is perfectly well-defined as an element of the $p$-adic rationals 
$$\mathbb{Q}_p = \text{Frac}(\mathbb{Z}_p) = \mathbb{Z}_p \left[ \frac{1}{p} \right]$$
even when $p = 2$; the $p$-adic rationals are a field of characteristic zero, and in any such field $x \mapsto {x \choose n}$ is well-defined. A $p$-adic rational has a $p$-adic valuation lying in $\mathbb{Z}$. 
