How to (dis)prove that for all prime numbers $p$, $p$ divides the binomial coefficient $\binom pk$ How should I give a proof or a counterexample for this statement. What I had thus far was $$\frac{p! \over {k!(n-k)!}}{p}$$
and thats not much since it’s clearly stated in the assignment itself.
I have no idea of how to prove or disprove the rest since they are all variables. I know that $p = rs$ with $r$ being $p$ and $s$ being $1$, but I have no clue about everything else. 
 A: ${p!\over {k!(p-k)!}}$ is an integer this implies that $k!(p-k)!$ divides $p!$ since $\gcd(p,k!(p-k)!)=1$ you deduce that $k!(p-k)!$ divides $(p-1)!$ and the result.
A: You can start from the following recurrence relation (which is used to prove the closed formula with factorials):
$$\forall k>0,\enspace\binom pk=\frac pk\binom{p-1}{k-1}\iff k\binom pk=p\binom{p-1}{k-1}.$$
This  shows $p$ divides the product $\; k\dbinom pk$. Now if $k<p$, $\;p$ doesn't divide $k$. As it is a prime, by Euclid's lemma, it divides $\;\dbinom pk$.
A: Since $\displaystyle \binom{p}{p} = \binom{p}{0}=1$, you must mean 
$\displaystyle p \left| \binom{p}{k} \right.$ for all $0 < k < p$.
We know that $\displaystyle \binom pk = \dfrac{p!}{k! (p-k)!}$ is an integer. Note that, since $0 < k < p$, then $0 < p-k < p$.  Hence $p \mid p!$ and, since $p$ is a prime number, $p \not \mid k!$ and $p \not \mid (p-k)!$
It follows that $\displaystyle p \left| \binom{p}{k} \right.$.
Even more simply, write out $\dfrac{p!}{k! (p-k)!}$ as
\begin{array}{cccccccccccccc}
1 & 2 & 3 & \cdots & k & (k+1) & (k+2) & \cdots & p\\
\hline
1 & 2 & 3 & \cdots & k & 1 & 2 & \cdots &(p-k)\\
\end{array}
Note that there is a $p$ in the numerator and not in the denominator. Since $p$ is prime, none of the factors in the denominator can multiply to cancel out or reduce the $p$ in the numerator. Hence $\displaystyle p \left| \binom{p}{k} \right.$.
