I'm quite lost on how to prove things, with the $n \choose k$ and proving. So the question is:
Prove that $n \choose k$ is divisible by $n$ if $n$ is a prime number and $1 \le k\le n-1$
Like, how would I/you go about proving that $n \choose k$ is divisible and is a prime number??
Thanks
On
A simple proof that uses Gauß's lemma. The binomial coefficients satisfy this recursion formula, for $0<k<n$:
$$\binom nk=\frac nk\binom{n-1}{k-1}$$ If $n$ is prime, $k$ is coprime with $n$, hence it divides $\dbinom{n-1}{k-1}$. This proves $\dbinom nk=n\times\text{some factor}$.
For the second question, if I've well understood it, $\dbinom nk$ is prime if and only if $n$ is prime and $k=1$ or $n-1$.
Indeed, note first $\dbinom nk=1$ if $k=0,n$ and $\dbinom nk=n$ if $k=1,n-1$. Secondly, if $2\le k\le n-2$, the relation above is a non-trivial factorisation of $\dbinom nk$, whence the assertion.
By definition, $\binom{n}{k}=\frac{n!}{k!(n-k)!}$. Also, since $k<n$ and $n$ is prime, then $k!$ and $(n-k)!$ both do not have a factor of $n$. Thus the quotient of $\frac{n!}{k!(n-k)!}$ is still divisible by $n$, as required.