Show: $(27!)^6 \equiv 1 \pmod{899}$. Note: $899=30^2 - 1$.
Could anyone show me how to go through this please? I think I have to use Wilson's theorem which is $(p-1)! \equiv -1 \pmod p$ but not exactly sure how else to tackle this. Thanks!
Show: $(27!)^6 \equiv 1 \pmod{899}$. Note: $899=30^2 - 1$.
Could anyone show me how to go through this please? I think I have to use Wilson's theorem which is $(p-1)! \equiv -1 \pmod p$ but not exactly sure how else to tackle this. Thanks!