Prove that, for prime $p$, the sum of products of numbers taken $r<p-1$ at a time from the set $\{1,2,\dots,p-1\}$ is always divisible by $p$.

Question

Prove that, for prime $p$, the sum of products of numbers taken $r<p-1$ at a time from the set $\{1,2,\dots,p-1\}$ is always divisible by $p$.

One way too prove is considering coefficients of polynomial: $$f(x)=(x-1)(x-2)...(x-p+1)-x^{p-1}+1.$$ Any other proof will be appreciated. The above proof of this on Introduction to Analytic Number Theory by Tom M. Apostol (Theorem 5.23).

Answer 1

Hint. Let $S_0=n$ and for $1\leq r\leq n$, $$S_r=\sum_{1\leq i_1<i_2<\dots <i_r\leq n} i_1i_2\cdots i_r$$ then (see Newton's_identities), $$S_r=\frac{1}{r}\sum_{j=1}^r (-1)^{j-1}S_{r-j}\sum_{k=1}^n k^j.$$ Now take $n=p-1$ and note that $\sum_{k=1}^{p-1} k^r$ is divisible by $p$ if $1\leq r<p-1$ (see for example Divisibility of a power sum by a prime).