This is a step in every proof finding the fixed field for the automorphism of $k(t)$ given by $t\mapsto t+1$ where $k$ is a field of characteristic $p$. I really don't know where to begin, but thought I would tackle a simpler case of simply trying to multiply out the following polynomial over $\mathbb{F}_p$:
$$\prod_{i=0}^{p-1}(x+i)=x^{p} + x^{p-1}\sum_{i-0}^{p-1}i + x^{p-2}\sum_{0\leq i<j\leq p-1}ij+\cdots$$
From here I have that $\sum_{i=0}^{p-1}i^2=(p-1)(p)/2\equiv0$. And similarly we have $$\sum_{0\leq i<j\leq p-1}ij=\sum_{j=2}^{p-1}\left(j\sum_{i=0}^{j-1}i\right)=\sum_{j=2}^{p-1}j\cdot j(j-1)/2=\frac{1}{2}\sum_{j=0}^{p-1}(j^3-j^2)$$
I can show that $\sum_{j=0}^{p-1}j^2 = \frac{1}{6} (p-1)p(2(p-1)+1)\equiv 0\mod p$ (if we assume $p\neq 3$). I feel this must not be the right track as it is already getting rather complicated, and I have no clear way of terminating this process.
Man, minutes after posting the question it finally hits me like a ton of bricks. $\mathbb{F}_{p}$ splits $x^p-x$, which is monic. So $x^p-x=\prod_{i=0}^{p-1}(x-i)$. And this immediately gives us the other result too.