Divisibility of $\binom{n}{p}$ by $p$, with $p$ prime A theorem of E. Lucas asserts that, if $p$ is a prime and $n,k$ are natural numbers, then :

$$\binom{n}{k}\equiv\binom{\lfloor n/p\rfloor}{\lfloor k/p\rfloor}\binom{n\text{ mod }p}{k\text{ mod }p}\pmod{p}$$

As a special case (replacing $k$ with $p$), we can see that :
$$\binom{n}{p}\equiv\lfloor\frac np\rfloor\pmod{p}$$
And, in particular :
$$p\mid\binom np\Leftrightarrow p\mid\lfloor\frac np\rfloor$$
First question : is there a way to prove that last property in some direct manner (ie : without having to know the general case) ?
Second question : I've found a numerical evidence that de $p$-adic valuation of $\binom np$ and $\lfloor\frac np\rfloor$ are equal (for $p$ prime). Is it true and how to prove it ?
 A: One way to approach your problem is the following: 
Recall $\cfrac{\binom{n}{p}}{\binom{n-1}{p}} = \cfrac{n}{n-p}$.
Let $\nu_p(n)$ denote the exponent of the highest power of $p$ dividing $n$.
Note that $n \equiv n-p$ mod $p$. 
You can now easily prove your result: 
$\nu_p\left(\binom{n}{p}\right) = \nu_p\left(\lfloor \cfrac{n}{p}\rfloor\right)$ for all prime $p$, for all $n \in \mathbb{N}$ by induction on $n$ using the following.
Take the base case $n=p$, the result is clearly true. 
A: Finally, I answer to myself ...
We know that :
$$p!\binom np=n(n-1)\ldots(n-p+1)$$
Let $i$ be the sole integer in $\{0,\ldots,p-1\}$ such that $p\mid n-i$. We can see that :
$$\frac{n-i}p\in\mathbb{N}\quad\text{and}\quad\frac{n-i}p\leqslant\frac np<\frac{n-i+p}p$$
so that :
$$\frac{n-i}p=\lfloor\frac np\rfloor$$
Let :
$$A=\prod_{0\leqslant j<p,\,j\neq i}(n-j)$$
The first equality can now be rewritten as :

$$(p-1)!\binom np=A\lfloor\frac np\rfloor$$

and since neither $(p-1)!$ nor $A$ is divisible by $p$, we conclude that $\binom np$ and $\lfloor\frac np\rfloor$ share the same $p-$adic valuation. In particular :
$$p\mid\binom np\Leftrightarrow p\mid\lfloor\frac np\rfloor$$
