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\item  Since $n$ is composite, we cknow $n$ can be factored in two cases.  Case 1: $n=a*b$ where $1 < a < b <(n-1)$ $\Rightarrow \frac{(n-1)!}{n} = \frac{1*2...a...b...(n-1}{n} \frac{1*2...a...b...n-1}{n}  = \frac{1*2...a...b...(n-1}{a*b}$ \frac{1*2...a...b...n-1}{a*b}$  Clearly, $n$ can divide $(n-1)!$. 
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Case 2: $n = a*a$ where $1 < a < 2a < n - 1$.   We know that $2a$ is less than $n-1$ since $n=a^2$, so as long as $n>2$, $n>2$,then  $2a < a^2$. $\frac{(n-1)!}{n} = \frac{1*2...a...2a...n-1}{n} = \frac{1*2...a...2a...n-1}{a*a}$     Clearly, $n$ can divide $(n-1)!$, so it always be $0\ \textrm{mod}\ n$ instead of $-1\ \textrm{mod}\ n$  Example: