So I'm trying to prove that for any natural number $1\leq k<p$, that $p$ prime divides:
$$\binom{p-1}{k} + \binom{p-2}{k-1} + \cdots +\binom{p-k}{1} + 1$$
Writing these choice functions in factorial form, I obtain:
$$\frac{(p-1)!}{k!(p-(k+1))!} + \frac{(p-2)!}{(k-1)!(p-(k+1))!} + \cdots + \frac{(p-k)!}{1!(p-(k+1))!} + 1$$
Thus as you can see each term except the last has a $(p-(k+1))!$ factor in its denominator. I've tried some stuff with integer partitions, tried to do some factoring and simplification, etc. But I can't see how to prove that $p$ divides this expression. I'll probably have to use the fact that $p$ is prime somehow, but I'm not sure how. Can anyone help me? Thanks.
Edit: Proof by induction is also a possibility I suppose, but that approach seems awfully complex since changing k changes every term but the last.