Question 104236
The factorial n! is defined for a positive integer n as


{{{n!}}} = {{{n(n-1)(n-2)}}}………{{{2*1}}}
as you can see, the last non-zero digit is {{{1}}}.

so
{{{96!}}} = {{{96*95*94}}}....{{{2*1}}} 

the last non-zero digit is {{{1}}}