If $p$ is a prime number, in Pascal's triangle all the terms in the $p$th row - except the 1s - are multiples of $p$ . It's easy to prove this property using the formula for $\binom{p}{k}$.
Is there an alternative proof, not using the formula for $\binom{p}{k}$?
Thanks.
The (cyclic) group with $p$ elements acts on subsets of $[p]=\{1,2,\ldots,p\}$ by cyclic permutations. The only two fixed points are the empty set and the full set $[n]$. All other orbits have $p$ elements, and are made up of subsets of one same size. Therefore, for any $k$ with $0<k<p$, the set $\binom{[p]}k$ of subsets of $[p]$ of size $k$ is a disjoint union of orbits each of size $k$, which shows that $p$ divides the number $\binom pk$ of such subsets.
This proof extends to proving $\binom{p^n}k\equiv0\pmod p$ for any $n\in\Bbb n$ and $0<k<p^n$, using the cyclic group of order $p^n$; there are more orbit sizes possible here, but only size$~1$ is not divisible by$~p$. More generally even there is Lucas's theorem.