Trying to prove that $p$ prime divides $\binom{p-1}{k} + \binom{p-2}{k-1} + \cdots +\binom{p-k}{1} + 1$ 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.

Letting $k=p-1$, we find that the expression equals $p$ summands of value $1$ each.

Maybe the problem was not meant to read "for some natural" but rather "for all natural" numbers $1\le k<p$.
The statement remains true, once you observer that the sum is simply $p\choose k$.
To see this combinatorially, note that you can choose $k$ out of $p$ objects by selecting $r$ with $0\le r \le k$, then take the first $r$ objects, do not take the next object and choose $k-r$ out of the remaing $p-1-r$ objects.
Finally, $p\choose k$ with $0<k<p$ is a multiple of $p$, for example because $p$ divides $p!$ but divides neither $k!$ nor $(p-k)!$. (This is where we need that $p$ is prime),
