deletions | additions       diff --git a/p7.tex b/p7.tex  index 02241fe..53cd8b3 100644  --- a/p7.tex  +++ b/p7.tex 
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\item  Let n=5:  $4! \equiv -1\ (\textrm{mod}\ n)$ \textrm{mod}\ n$  $4! \equiv -1\ (\textrm{mod}\ 5)$ \textrm{mod}\ 5$  $4*3*2*1 \equiv -1\ (\textrm{mod}\ 5)$ \textrm{mod}\ 5$  $(4*1)(3*2) \equiv -1\ (\textrm{mod}\ 5)$ \textrm{mod}\ 5$  $(4*1) \equiv -1\ (\textrm{mod}\ 5)$ \textrm{mod}\ 5$  $(3*2) \equiv 1\ (\textrm{mod}\ 5)$ \textrm{mod}\ 5$  Then via Wilson's Theorem, it will then equal to $1\ \textrm{mod}\ 5$  \end{itemize}