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.
Best Answer
$$\binom{p-1}{k} + \binom{p-2}{k-1} + \cdots +\binom{p-k}{1} + 1= \sum_{i=0}^k\binom{p-k-1+i}i =\binom pk, $$ and it is well known that a prime $p$ divides all binomial coefficients $\binom pk$ with $k\notin\{0,p\}$ (cf. the Frobenius endomorphism of rings of characteristic $p$). The fact that $\sum_{i=0}^k\binom{n+i}i=\binom{n+k+1}k$ is immediate by recurrence on $k$ from the Pascal's rule, the basic recurrence defining binomial coefficients.
If you prefer combinatorial arguments to algebra: you can understand the summation identity by classifying the $k$-combinations of a $p$-set according to the first element that is not in the $k$-combination (if the very first element of the $p$-set is not in the $k$-combination, then one has a $k$-combination of the remaining $p-1$; if the first element is in the $k$-combination but the second one is not, then $k-1$ of the remaining $p-2$ elements must be chosen to complete the $k$-combination; and so forth, until element $k+1$ which cannot be in the $k$-combination if all values before it are already in, contributing $1$), and you can understand the divisibility $p\mid\binom pk$ by the fact that by cyclically rotating $k$-combinations, one can group them into orbits of $p$ distinct $k$-combinations each.