$$ \sum_{d \mid (n,k_1,k_2, \dots,k_m)}\mu(d)\binom{n/d}{k_1/d, k_2/d, \dots, k_m/d} \equiv 0 \pmod n $$ where $\mu$ is the Moebius mu function. I've found above interesting divisibility properties. I've already proven this and it seems just exercise-level.
I would like to know it is alreay proven in somewhere.