TITLE: Use Wilson theorem to show that $63! + 1 \equiv 0 \mod ~ 71$
QUESTION [5 upvotes]: Use Wilson theorem to show that $63! + 1 \equiv 0 \mod ~ 71$.

71 is prime then Wilson theorem says that $(71-1)!+1=0 \mod ~ 71$ i.e $70!+1\equiv 0 \mod ~ 71$ then how to proceed further?

REPLY [3 votes]: You have to show that $70! \equiv 63! \mod 71$, so it suffices to show that $70 \cdot 69\cdot 68\cdot 67\cdot 66 \cdot 65 \cdot 64 \equiv 1 \mod 71$. This holds because of $70 \cdot 69\cdot 68\cdot 67\cdot 66 \cdot 65 \cdot 64 \equiv (-1)\cdot (-2)\cdot (-3)\cdot (-4)\cdot (-5)\cdot (-6)\cdot (-7) \equiv -720 \cdot 7 \equiv -10 \cdot 7 \equiv -70 \equiv 1 \mod 71$