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 us define  $jp$, with $j$ positive integer, the greatest multiple of $p$ lower than $k$. Now let us consider the definition of  $\binom{k}{p}$ as $$\frac {k (k-1)(k-2)(k-3)...[k-(p-1)]}{p!}$$ The numerator is formed by $p$ consecutive numbers and can be expressed as a product of factors  that are $\equiv i \pmod p$, where $i$ assumes all integer values between $0$ (this is the case of $jp$) and $p-1$. Because $1 \cdot 2 \cdot 3....\cdot (p-1) \equiv -1 \pmod p$, the whole numerator can be written as the product of $jp$ and a number $\equiv -1 \pmod p$. As a result, dividing the numerator by $p $ we get a number that is $\equiv -j \pmod p$. 
On the other hand, the denominator $p!=p(p-1)!$ is the product of $p$ and a quantity $\equiv -1 \pmod p$, so that dividing it by $p$  we get a number that is $\equiv -1 \pmod p$. 
Based on the considerations above, $\binom{k}{p}$ is given by a fraction where, simplifying both numerator and denominator by $p$, we are left with a ratio between a number that is $ \equiv -j \pmod p $ and another one that is $ \equiv -1 \pmod p $. It follows that the result of the fraction is $ \equiv j \pmod p $. Because $j=\lfloor k/p \rfloor $, we finally obtain
$$\binom{k}{p} \equiv \left\lfloor \frac{k}{p} \right\rfloor \pmod p$$
EDIT: since a comment asked to better explain the calculations for the numerator, I added an example below.
Let us set, for example, $k=7$ and $p=3$. So we have $jp=6$ and $j=2$. The binomial coefficient is given by the ratio
$$\frac {7 (7-1)(7-2)}{3!} $$
The numerator is formed by $3$ consecutive numbers and can be expressed as a product of factors  that are $\equiv i \pmod 3$, where $i$ assumes all integer values between $0$ and $p-1=2$. In fact, the numerator can be written as $6(6+1)(3+2) $. The product of these factors, not considering $jp=6$, is $$(p-1)! \mod p = (3-1)! \pmod 3=2 \pmod 3$$
 (note that we can also consider this last number as $-1 \pmod p $, although it is not necessary).  Then, the numerator can be written as the product of $jp=6$ and a number $\equiv 2 \pmod 3$, i.e. it is $\equiv jp (p-1)! \pmod p=\equiv (6 \cdot 2) \pmod 3$. 
The denominator is $3! $ and then can be written as the product of $p=3$ and $(p-1)!=2$, where this last number is $\equiv (p-1)! \pmod p = \equiv 2 \pmod 3$.
Then the initial binomial coefficient is the ratio of a quantity $ \equiv  (6 \cdot 2) \pmod 3$ and another quantity that is $ \equiv (3 \cdot 2) \pmod 3$. Their ratio is then a number $\equiv 2 \pmod 3$, where the number $2$ simply reflects the initial value of $j$, which in turn is equal to $   \left\lfloor \frac{7}{3} \right\rfloor$.  We then conclude that 
$$\binom{7}{3} \equiv \left\lfloor \frac{7}{3} \right\rfloor \pmod 3$$
