4
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By Wilson's theorem we know that if $n$ is a prime number then $(n-1)! \equiv n-1 \pmod n$

So, upon division by $n-1$ on both the sides we have $(n-2)! \equiv 1 \pmod n$

Edit 1: The teacher deducted marks in this proof and said that I've to consider the case for mod 2 separately. Also, her claim is verified by proof given here .

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    $\begingroup$ Why do you think there is a fallacy? Since $n-1$ is relative prime to $n$, you can divide like this. $\endgroup$ – Thomas Andrews Oct 18 '16 at 17:28
  • $\begingroup$ en.wikipedia.org/wiki/Wilson%27s_theorem#Proofs that's not a fallacy; see the wikipedia article, in particular, "Proof Using the Sylow Theorems" $\endgroup$ – Giuseppe Oct 18 '16 at 17:29
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    $\begingroup$ Perhaps it clarifies things for you if you write $(n-1)!\equiv -1\pmod n$ and remark that $n-1\equiv -1\pmod n$. $\endgroup$ – lulu Oct 18 '16 at 17:29
  • $\begingroup$ The teacher deducated marks in this proof and said that I've to consider the case for mod 2 separately. Also, her claim is verified by proof given at brilliant.org/wiki/wilsons-theorem/…" $\endgroup$ – ankit Oct 18 '16 at 17:32
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    $\begingroup$ Unless the teacher has issues with $0!$, there is no real reason to treat $2$ differently. $\endgroup$ – Daniel Fischer Oct 18 '16 at 17:43

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