Proving ${n \choose p} \equiv \Bigl[\frac{n}{p}\Bigr] \ (\text{mod} \ p)$ This is an exercise from Apostol, which i have been struggling for a while.
Given a prime $p$, how does one show that $${n \choose p} \equiv \biggl[\frac{n}{p}\biggr] \ (\text{mod} \ p)$$ Note that $\Bigl[\frac{n}{p}\Bigr]$ denotes the integral part of $\frac{n}{p}$.
I would also like to know as to how does one try to solve this problem. Well, what we need is to show is whenever one divides ${n \choose p}$ by a prime $p$ the remainder is the integral part of $\frac{n}{p}$.
Now, $${ n \choose p} = \frac{n!}{p! \cdot (n-p)!}$$ Now $n!$ can be written as $$n!= n \cdot (n-1) \cdot (n-2) \cdots (n-p) \cdots 2 \cdot 1$$
But i am really struggling in getting the integral part.

You can see the solution for the case $p=7$ here: http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1775313.
