Jeremy Ting edited p7.tex  about 10 years ago

Commit id: 17162df76eba74eef6542825580e872eadeeea5e

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Let n=5:  $4! \equiv -1\ (\textrm{mod}\ n)$  $4! \equiv -1\ (\textrm{mod}\ 5)$     $4*3*2*1 \equiv -1\ (\textrm{mod}\ 5)$     $(4*1)(3*2) \equiv -1\ (\textrm{mod}\ 5)$     $(4*1) \equiv -1\ (\textrm{mod}\ 5)$     $(3*2) \equiv 1\ (\textrm{mod}\ 5)$     Then via Wilson's Theorem, it will then equal to $1\ (\textrm{mod}\ 5)$  \end{itemize}