Prove that $\binom{k}{p} \equiv \left\lfloor \frac{k}{p} \right\rfloor \pmod p$ for all odd prime number $p$ and $k\ge 3$ I don't really know how to start this exercise. Do I have to use p-adic valuation ?
If it's the case it will give $\nu_p(\binom{k}{p})= \sum \limits_{r=1}^{\infty}\left(\left\lfloor \frac{k}{p^r} \right\rfloor-\left\lfloor \frac{p}{p^r} \right\rfloor -\left\lfloor \frac{k-p}{p^r} \right\rfloor \right)$.
Thanks in advance !

show that :


*

*${k \choose p} = \frac{k}{k-p}{k-1 \choose p} = (1+\frac{p}{k-p}){k-1 \choose p}$
so if $k \not\equiv 0 \bmod p$ then $k-p \ | {k-1 \choose p}$  and ${k \choose p} \equiv {k-1 \choose p} \bmod p$.

*${ap \choose p} \equiv a \bmod p$, since ${ap \choose p} = a {ap-1 \choose p-1}$ and ${ap-1 \choose p-1} = \prod_{m=1}^{p-1} \frac{ap-m}{m} \equiv \left(\prod_{m=1}^{p-1} (p-m) \right)\left(\prod_{m=1}^{p-1}m\right)^{-1}\equiv 1 \bmod p$ 
