If $p$ is a prime integer, prove that $p$ is a divisor of $\binom p i$ for $0 < i < p$ I was thinking of using the definition for combinations and use the fact that $p$ appears in the expansion of $\binom pi$ and hence $p$ is a divisor. I don't know whether I am on the right track!
 A: We have $$\binom{p}{i} = \dfrac{p!}{i!(p-i)!} = \dfrac{p(p-1)\cdots(p-i+1)}{i!}$$
Now notice that $i!\binom{p}{i} = p(p-1)\cdots(p-i+1) \implies   p|i! $ or $p|\binom{p}{i}$.
But $p|i!$ is not possible for $0\lt i\lt p$. That means $p|\binom{p}{i}$.
A: I propose you a different combinatoric proof. Fix $n \in \{1,\ldots,p-1\}$.
Let $1\le x_1<\ldots<x_n\le p$ be some integer, and notice that they have different remainders when they are divided by $p$. Obviously also
$$x_1+k,x_2+k,\ldots,x_n+k$$
have different remainders when they are divided by $p$, for each integer $k \in \{0,1,\ldots, p-1\}$. Moreover, all the remainders are "shifted together", meaning that if we put $k=p$ we get the original set of residues.
Therefore, we are establishing a relation between subsets of $n$ elements $\{1,2,\ldots,p\}$ which is reflexive, symmetric and transitive: it is an equivalence relation. 
Let's consider an equivalence class now: it is made by exactly $p$ elements. It implies that the total number of subsets, which is $\binom{p}{n}$, has to be a multiple of $p$.
A: First we give a direct proof; then we explain how to view the proof in denominator order language, i.e. that a fraction is writable with coprime denominators iff it is an integer.
$\ \displaystyle {p \choose i}\ =\, \color{#0a0}{\dfrac{p!}{(p\!-\!i)!}}\dfrac{1}{i!}\, =\, \dfrac{\color{#0a0}{p(p\!-\!1)\cdots (p\!-\!i\!+\!1)}}{i\, (i\!-\!1)\cdots 2\cdot 1}\, =\, \dfrac{pa}b\in \Bbb Z.\ $
By Euclid's Lemma $\ \ \ \color{#c00}{p\nmid b}\ $ and $\ \color{#c00}{p\mid b}\left(\dfrac{pa}{b}\right) \Rightarrow\ \displaystyle p\ {\large  \mid}\ \dfrac{pa}b  $
In this case we have  $\ p\nmid b = b_i\cdots b_2 b_1\,$ else, by $\,p\,$ prime, $\,p\mid b_i\,$ for some $\,i,\,$ contra  $\, 0 < b_i < p$
Alternatively note $\,\dfrac{pa}{b} = c\in\Bbb Z\,\Rightarrow\, \dfrac{a}b = \dfrac{c}p\,$ is a fraction writable with coprime denominators. Therefore, by the Lemma below, $\,a/b = \color{#b0f}{c/p\in \Bbb Z},\,$ hence $\,\color{#b0f}{p\mid c} = pa/b,\,$ as claimed.
Lemma $\ $ A fraction can be written with coprime denominators iff it is an integer.
Proof $\ $ If $\, q = a/b = c/d\,$ for coprime $\,b,d\,$ then $\, \color{#c00}{jb\!+\!kd = 1}\,$ for $\,j,k\in\Bbb Z\,$ by Bezout, so
$$ bq,dq\in\Bbb Z\,\Rightarrow\, j(bq)+k(dq) = (\color{#c00}{jb\!+\!kd})q = q\in \Bbb Z\qquad\qquad\qquad$$
The Lemma is simply a denominator form of the following ubiquitous group theory theorem:
If $\,q^b\! = 1 = q^d$ for coprime $\,b,d\,$ then $\,q = 1,\ $ by $\ n={\rm ord}(q)\,$ divides coprime $\,b,d\,$ so $\,n = 1.\,$
The least denominator of a fraction is its order in $\,\Bbb Q/\Bbb Z,\,$ so the above is a special case of this result. For  more on this viewpoint (denominator and order ideals) see here.
A: Hint: recall definition $\binom{p}{i}=\frac{p!}{i!(p-i)!}$ and note that $p$ divides the numerator, but not the denominator.
A: Hint:-
$\gcd(p,i)=\gcd(p,p-i)=1$ for all $0<i<p$.
