If $p$ be a prime and r be any integer, $0 < r < p$ then $\frac{(p-1)!}{r!(p-r)!}$ is an integer. 
Let $p$ be a prime and $r$ an integer, such that $0 < r < p$. Show that $\frac{(p-1)!}{r!(p-r)!}$ is an integer.


The given number is $ \displaystyle\frac{1}{p}\binom {p} {r}$. after tha how can I show that$p$ divides the given binomial coefficient. can somebody help me please.
 A: Write
$$p!=r! \cdot (p-r)! \cdot \binom{p}{r}$$
Since $p|p!$ then $p| r! \cdot (p-r)! \cdot \binom{p}{r}$. Using the fact that $p$ is prime you get
$$p|r! \, \mbox{or} \, p| (p-r)! \, \mbox{or} \, p|\binom{p}{r}$$
Now, the first two options cannot be true, thus....
A: Look at the factors in the denominator - which of these can have a common factor with $p$? - Use that $p$ is prime and you have a factorisation of the denominator into integers less than $p$.
A: Let your number be $q$.
Since $\gcd(r,p) = 1$, we have that there are integers $a,b$ such that $ar + bp = 1$.
Muliplying by $q$, and we get
$$a (qr) + b (pq) = q$$
Now $qr = \binom{p-1}{r-1}$ and $pq = \binom{p}{r}$, thus $q$ is an integer.
Another method is to use the fact that
$$(1+x)^p = 1 + x$$
in $F_p[x]$
A: Let $p$ be a prime and let $0<r<p$. Then $b(p,r)=\binom{p}{r}$ is an integer, because it is a binomial coefficient (which appears as a coefficient of $x^r$ in $(1+x)^p$). Moreover,
$$b(p,r)=\binom{p}{r} = \frac{p!}{r!(p-r)!} = \frac{p \cdot (p-1)!}{r!(p-r)!}.$$
Since $0<r<p$, neither $r!$ nor $(p-r)!$ have a factor of $p$. Hence $p$ only appears in the numerator of the expression for $b(p,r)$, and it follows that the integer $b(p,r)$ is a multiple of $p$ or, in other words, the expression
$$\frac{b(p,r)}{p}=\frac{(p-1)!}{r!(p-r)!}$$
is also an integer. 
PS: The formal justification of the last step is as follows. The identity of integers
$$b(p,r) \cdot r!\cdot (p-r)! = p \cdot (p-1)!$$
implies that $p$ divides the left-hand side $b(p,r)\cdot (r!(p-r)!)$. Since $p$ is prime, it divides one of the two factors: $b(p,r)$ or $r!(p-r)!$. But $0<r<p$ implies that $p$ does not divide $r!$ nor $(p-r)!$, and so $p$ is a divisor of $b(p,r)$. This means that $b(p,r)/p$ is an integer. 
A: Using this, $r! (p-r)!$ divides $p!$ for $1\le r<p$
$\implies r! (p-r)!$ divides $(p-1)!$ as $(p,r)=1$ for $1\le r<p$
