Let $p$ be a prime. Show that $${n \choose p}-\bigg[\dfrac{n}{p}\bigg]$$ is divisible by $p$, for all $n\in \mathbb{N}$.
I could not attempt this problem at all. Please help. Thank you.
EDIT: Here $\bigg[ \cdot \bigg]$ is the greatest integer function.
This is trivial if $n \leq p$, so assume $n > p$.
Note the relationship
$\binom{n}{p} \cdot p! = n \cdot (n-1) \cdots (n-p+1)$
The product on the right side is a list of p consecutive integers, so exactly one of them is divisible by $p$. In fact, that number is
$\left\lfloor \frac{n}{p} \right\rfloor p$.
The product of the remaining p-1 factors on the right side is congruent to $(p-1)!$. Thus, dividing both sides by $p$ and then reducing modulo $p$ gives
$\binom{n}{p} \cdot (p-1)! \equiv \left\lfloor \frac{n}{p} \right\rfloor (p-1)! \pmod{p}$
Now cancel the $(p-1)!$ (allowed since it is not divisible by $p$) and the result follows.