For all $n \in \mathbb{N}$ and prime $p$ i was asked to prove that $p$ divides $\binom{n}{p}$ -$[n/p]$. Where $[\text{...}]$ denotes box function and $\binom{n}{p}$ denotes $n$ choose $p$.
Can not understand where to start from. Is it possible by congruency? Please help.
By Lucas's theorem $\binom{n}{p}$ is congruent to $\binom{w}{1}$ where $w$ is the second digit in the base $p$ expansion of $b$ ( because the base $p$ expansion of $p$ is just $10$). So $\binom{n}{p}$ is congruent to $w$ and $w$ is clearly congruent to $\lfloor n/p\rfloor$