Prove that $\frac{\binom{p}{k}}{p}$ is integral for $k\in \{1,..,p-1\}$ with $p$ a prime number with a particular method I started by induction on $k$ 
For $k=1$ then : $1\in \mathbb{N}$
For $k=2$ then : $\frac{(p-1)}{2!} \in \mathbb{N}$ , indeed for all $p>{2}$, $p-1$ is even. (We still have $k<p$ it's important).
Suppose that $\frac{(p-1)...(p-k+1)}{k!} \in \mathbb{N}$ for a specific $k \in \{1,..,p-1\}$ 
Now try to prove that : $\frac{(p-1)...(p-k)}{(k+1)!} \in \mathbb{N}$.
By the recursive formula we have directly :
$\frac{\binom{p}{k+1}}{p} = \frac{[\binom{p+1}{k+1} - \binom{p}{k}]}{p}$ $\Leftrightarrow$ $\frac{(p-1)...(p-k)}{(k+1)!} = \frac{(p+1)(p-1)...(p-k+1)}{(k+1)!} - \frac{(p-1)...(p-k+1)}{k!}$
By hypothesis we know that $\frac{(p-1)...(p-k+1)}{k!} \in \mathbb{N}$
Now, the key is to see that a product of $k$ consecutive terms is divided by $k!$ (a little induction can solve this).
Then the induction is complete.

By definition,
$$\binom{p}{k} = \frac{p!}{k!(p-k)!}$$
Since $p$ is prime, it's relative prime to all smaller numbers, and therefore for $k<p$, it cannot be a factor of $k!$. Now if further $k>0$, then $p-k<p$, and thus $p$ is not a factor of $k!(n-k)!$. Since by definition, $p$ is a factor of $p!$, and $p$ is prime, it follows that it is a factor of $\binom{p}{k}$, thus $\binom{p}{k}/p\in\mathbb N$.
