# Conditions for which $n | {n \choose k}$ for all $k$

I'm studying for a number theory exam. Our review sheets offers the question:

Under what conditions will $n$ divide $n \choose k$ for all 1 $\leq k \leq n-1$?

I can see that this will be true for any prime $n$, and don't think that it would be true for any composite $n$, but am unsure in what direction I should proceed with the proof (prime decomposition, congruence, etc). Any and all help will be greatly appreciated!

With that in mind, here is a suggestion that comes from playing around with binomial coefficients. If $n$ is composite, then clearly $n\choose 0$, $n\choose 1$, $n\choose {n-1}$, and $n\choose n$ cannot provide counterexamples. But there is one very good number to check, the simplest one that makes use of the fact that $n$ is composite: $n\choose p$, where $p$ is the smallest prime factor of $n$.
(Some of this is mentioned in the link in the comments, but I feel that most of the answers there overcomplicate things just a tad. Still it's good to be aware of the general patterns, so you can check out the link for a more general discussion about what happens for other factors and non-factors of $n$.)