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 !

I try to explain better my previous comment on induction's argument:
let's suppose $\binom{k}{p}=n$, so that $np\leq k <np+p$;
it's easy to prove $\binom{k+1}{p}=\frac{k+1}{k+1-p}\binom{k}{p}\equiv 1\left\lfloor \frac{k}{p} \right\rfloor\pmod{p}$, 
if $k+1\not\equiv 0\pmod p$, and in this case we are done since $k+1<pn+p$ and so $\left\lfloor \frac{k+1}{p} \right\rfloor=\left\lfloor \frac{k}{p} \right\rfloor$.
If otherwise $k+1=p^{r}a,a\not\equiv 0\pmod p$ then $\left\lfloor \frac{k+1}{p} \right\rfloor=p^{r-1}a$ and  $\binom{k+1}{p}=\frac{p^{r-1}a}{p^{r-1}a-1}\binom{k}{p}$; 
Now if $r\geq 2$ then  $\binom{k+1}{p}\equiv 0\pmod p$ and similarly for  $\left\lfloor \frac{k+1}{p} \right\rfloor=p^{r-1}a$;
if $r=1$ then  $\binom{k+1}{p}=\frac{a}{a-1}\binom{k}{p}$ and its easy to see that $a=n+1$ so that  $\binom{k+1}{p}=\frac{n+1}{n}\binom{k}{p}$;
But now if $n\not \equiv 0 \pmod p$ since $n=\left\lfloor \frac{k}{p} \right\rfloor$ and $n+1=\left\lfloor \frac{k+1}{p} \right\rfloor$ we get $\binom{k+1}{p}=\frac{n+1}{n}\binom{k}{p}\equiv \left\lfloor \frac{k+1}{p} \right\rfloor\pmod p$.
Finally if $n=p^{c}m$ with $m\not\equiv 0\pmod p$ then $\left\lfloor \frac{k+1}{p} \right\rfloor\equiv 1\pmod p$ and $\binom{k}{p}=\binom{p^{c+1}m+p-1}{p}=\frac{(p^{c+1}m+p-1)...(p^{c+1}m)}{p(p-1)!}=p^{c}mw$, with $w\equiv 1\pmod p$
and we get the desired equivalence, since $\binom{k+1}{p}=\frac{p^{c}m+1}{p^{c}m}\binom{k}{p}=(p^{c}m+1)w\equiv 1\pmod p$.
