Past exam of the mathematics course of the University of Cambridge / 2011 / ii / Paper 3 / 1I / ii / Solution

Put $m=(p-1)/2$ and let $O=1\cdot 3\cdot 5\cdots (p-2)$. The even residues $2,4,\dots ,p-1$ are precisely the negatives, in reverse order, of the odd residues $1,3,\dots ,p-2$. Hence their product is $(-1{)}^{m}O$ in the finite field ${\mathbb{F}}_p$. Applying Wilson theorem givesThe initial factor $1^2$ can be omitted, soFor $p=3$ the product is empty and equals $1$, consistently with the formula.