p-adic numbers:
Consider the ring of p-adic integers $ \ \mathbb{Z}_p $.
Show that if $ \ a \in \mathbb{Z}_p \ $, then $ \ \frac{a(a-1) \cdots (a-n+1)}{n!} \in \mathbb{Z}_p$ also, where $n$ is natural number.
Answer:
To show that $\frac{a(a-1) \cdots (a-n+1)}{n!} \in \mathbb{Z}_p$, we have to show that $ \left|\frac{a(a-1) \cdots (a-n+1)}{n!} \right|_p \leq 1$.
I have seen some hintz in the book of "p-adic numbers, p-adic analysis and zet-function" of the autor "Neal Koblitz". The hintz is given below:
I did not understand how they have chosen $N$ and what does $ord_p(a-a_0)>N$ mean?
please help to understand this.
or
Any other method to show that $ \ a \in \mathbb{Z}_p \Rightarrow \frac{a(a-1) \cdots (a-n+1)}{n!} \in \mathbb{Z}_p$ .
Because of the denominator $n!$, the point is that $a(a-1)...(a-n+1)/n!$ lives a priori in $\mathbf Q_p$, not necessarily in $\mathbf Z_p$. Whatever your construction of $\mathbf Z_p$ may be, you must think that an element $a\in \mathbf Z_p$, by definition, is"$p$-adically approximated" at any order $N$ by an $a_0 \in \mathbf Z$. This locution means that for any integer $N$, there exists $a_0 \in \mathbf Z$ s.t. $ord_p (a-a_0)>N$, which means in turn (by definition), that in $\mathbf Z_p$, you can write $a-a_o=up^M$, where $u\in \mathbf Z^*_p$ and $M>N$. Consequently, given $n$, the fraction $a(a-1)...(a-n+1)/n!\in \mathbf Q_p$ can be approximated at any order by $a_0(a_0-1)...(a_0-n+1)/n!$ (since $n$ is fixed, just take $N$ large enough). The latter fraction lives in $\mathbf Z$ because it expresses the number of combinations of $a_0$ objects among $n$, so the given fraction lives in $\mathbf Z_p$ .
EDIT: In case the final argument is not clear enough (see the OP query), I develop it in detail. For $N$ large enough, the two fractions (involving resp. $a$ and $a_0$) are $p$-adically close. But since the $p$-adic valuation is discrete, this means that the two fractions have the same $p$-adic absolute value, and we are done.