Let $p$ be a prime number, and $n$ be an integer such that $n \geq p$. Let $a_1,...,a_n$ be arbitrary integers. Let $s_0 = 1$, and for every $k \ge 1$, let $$s_k=|\{B \subset \{1,2,...,n\} : p\mid\sum_{i \in B}a_i \text{ and }|B|=k\}|.$$ Show $$p\mid\sum_{k=0}^n(-1)^ks_k.$$
Attempt so far: $\sum_{k=0}^n(-1)^ks_k$ is the number of even subsets of $\{a_1,...,a_n\}$ that is divisible by $p$ minus the number of odd subsets that is divisible by $p$.
If we view the $a_i$s in mod $p$, then clearly, all single subset that is divisible by $p$ is $o$ mod $p$.
then the subset with $2$ elements whose sum is divisible by $p$ are inverse of each other under addition mod $p$