Proof verification ($p$ prime then $p\mid \binom{p}{k}$ for $1\le k\le p-1$)

Question

We have to prove that $$p\mid\binom{p}{k}$$ for $1\le k\le p-1$.

My approach was to expand $\binom{p}{k}$ and simplify as

$$\binom{p}{k}=\frac{1\cdot 2\cdot\ldots\cdot p}{1\cdot 2\cdot\ldots\cdot k\cdot 1\cdot 2\cdot\ldots\cdot (p-k-1)\cdot (p-k) }=$$

$$=\frac{(p-k+1)\cdot\ldots\cdot (p-1)\cdot p}{k!}=p\cdot\frac{(p-k+1)\cdot\ldots\cdot (p-1)}{k!}.$$

Now (because of combinatorial reasons) $\binom{p}{k}$ is an integer so the last expression above has to be an integer as well. This means that $\frac{(p-k+1)\cdot\ldots\cdot (p-1)}{k!}$ is eighter an integer self (we have then $\binom{p}{k}=p\cdot n, n\in\mathbb{Z}$) and then we are done or it becomes an integer after multiplication by $p$. Now we simplify in this expression so that the nominator and denominator is relative primes so we get

$$\frac{(p-k+1)\cdot\ldots\cdot (p-1)}{k!}=\frac{a}{b}$$

with $\gcd(a,b)=1$. Now $p$ wasn't a factor in $k!$ since $k\le p-1$ and $p$ is prime and after the simplification we only can have less factors in the denominator so $p$ cannot be one of the factors. This means that $p=b\cdot m, m\in\mathbb{Z}$ but since $p$ was prime this implies eighter $b=1$ or $b=p$. Now if $b=1$ then the expression is an integer and we are done. And $b=p$ means that the product of all the factors left after the simplification gives $p$ and since it was prime this means that one of the factors was $p$ itself which is a contradiction because we already saw that all the factors in the denominator has to be smaller than $p$. Q.E.D

This feels like a standard problem that one should tackle with induction but I still would like to know whether this is correct. Any help? Thanks.

Answer 1

$k\binom{p}{k} = (p-k+1) \binom{p}{k-1}$ and then you do an induction on $k$, thanks to the fact that $p-k+1$ is prime to $p$. You almost got it !

Variation : you have $$k {p\choose k}=p{{p-1}\choose {k-1}}$$ and since $p$ and $k$ are prime, then $p|{p\choose k}$.

Answer 2

Here's a proof without using induction.

C(p,k) = p! / k! (p-k)!. So C(p,k) k! (p-k)! = p!. Consider the unique factorization in prime numbers of C(p,k) k! (p-k)! = p!. The prime factors of C(p,k) k! (p-k)! are the prime factors of C(p,k) together with the prime factors of k! and the prime factors of (p-k)! The prime factor p must belong to C(p,k) because it does not belong to k! or (p-k)! as k< p and p is prime. Hence p divides C(p,k).