How do I show that $p\mid {p\choose k}$ if $gcd(p,k)=1$?

Question

Suppose we need to show that $p \mid {p\choose k}$ if $(p,k)=1$. Without using the fundamental theorem of arithmetic, is it possible to prove this?

I'm thinking of using the following steps:

1. By expressing $p\choose k$ as $\frac{p(p-1)!}{(p-k)!k!}$, we can show p that $p\choose k$ is a multiple of p, and the statement is immediately true.
2. But problem is I don't know if $\frac{(p-1)!}{(p-k)!k!}$ is a fraction, in which case I will be wrong.

Answer 1

Almost there! Use $$k{p\choose k} =p\frac{(p-1)!}{(p-k)!(k-1)!}$$ and the fact that the right-hand side is $p$ times another binomial coefficient.

Answer 2

I suppose you mean that $p$ is a prime number. Remind that if $p \wedge a=1$ and $p\wedge b=1$ then $p \wedge (ab)=1$. ( follows from Bezout's theorem )

Remark that $$ k!\binom{p}{k}=p\left(p-1\right) \dots \left(p-k+1\right) $$ Hence $$ p \ | \ k!\binom{p}{k} $$ Using that $p \wedge k=1$ for all $k \in [1,p]$ then $p \wedge k!=1$. Using Gauss's theorem

$$ p \ | \ \binom{p}{k} $$

Answer 3

Note that $$ \binom{p}{k}=\frac pk\binom{p-1}{k-1} $$ If $(p,k)=1$, Bezout's Identity says there are $x,y\in\mathbb{Z}$ so that $$ px+ky=1 $$ Then $$ \binom{p}{k}=p\left[y\binom{p-1}{k-1}+x\binom{p}{k}\right] $$