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Let $n \ge m \in \mathbb{N}$ . Let $a_0, b_0$ be the least non-negative residues of $n,m \pmod{p}$ , respectively. (Additionally, we set $r=n-m$ , and $r_0$ is the least non-negative residue of $r$ modulo $p$ .) Then the statement follows from
We define the 'carry indicators' $c_i$ for all $i \ge 0$ as
and additionally $c_{-1} =0$ .
The special case $s=1$ of Anton's congruence is: \begin{equation} \pfac{n} \equiv \left(- 1\right)^{\left\lfloor \frac{n}{p}\right\rfloor}\cdot a_0! \pmod{p}, \end{equation}where $a_0$ as defined above, and $\pfac{n}$ is the product of numbers $\le n$ not divisible by $p$ . So we have
When dividing by the left-hand terms of the congruences for $m$ and $r$ , we see that the power of $p$ is
So we get the congruence
or equivalently \begin{equation} \label{L1} \left(\frac{-1}{p}\right)^{c_0}\cdot \binom{n}{m} \equiv \binom{a_0}{b_0}\binom{\left\lfloor\frac{n}{p}\right\rfloor}{\left\lfloor\frac{m}{p}\right\rfloor} \pmod{p}. \end{equation} Now we consider $c_0 =1$ . Since $$ a_0 =b_0 +r_0 -\underbrace{pc_0=p} $$ $b_0 +r_0 < b_0 \leftrightarrow c_0 =1 \leftrightarrow b_0 -(p -r_0) =a_0 <b_0 \leftrightarrow \binom{a_0}{b_0}=0 $ . So both congruences-the one in the statement and (![[*]](http://images.planetmath.org:8080/cache/objects/3917/js//usr/share/latex2html/icons/crossref.png "[*]") )- produce the same results.
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