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 !

Let $P(x)=x(x-1)\cdots (x-p+1)$. We know that $P(x)=x^p-x+pS(x)$, with $S\in \mathbb{Z}[x]$. Hence we get that $P^{\prime}(x)=px^{p-1}-1+pS^{\prime}(x)$, hence $P^{\prime}(x)=-1\mod{p}$. Now we have $P(x+p)=P(x)+pP^{\prime}(x)\mod{p^2}$, hence $P(x+p)=P(x)-p\mod{p^2}$. Let now be $Q(x)=P(x)/p!$. By the above, we get that for $x\in \mathbb{Z}$, we have 
$$Q(x+p)=Q(x)-\frac{p}{p!} \mod{p}$$
But $-\frac{p}{p!}=-\frac{1}{(p-1)!}=1\mod{p}$.
Hence we get that for $k\in \mathbb{N}$, we have $$\binom{k+p}{p}=\binom{k}{p}+1\mod{p}$$ 
And now an induction finish the job. 
