By Wilson's theorem, I have that $(p-1)!\equiv-1\pmod p$ for any odd prime $p$ which means that $p\mid (p-1)!+1$
But how does it reduce to that the remainder when $(p-1)!$ is divided by $p$ is equal to $p-1$?
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$$p\mid -1+1=0 \Leftrightarrow p\mid (p-1)+1=p$$ Don't mix up "$\equiv$" and "$=$". "$a\equiv b \pmod p$" means $p\mid a-b$. – Wolfgang Kais Aug 08 '20 at 22:56
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Your last sentence is too ungrammatical to be understood. Please edit. – DanielWainfleet Aug 10 '20 at 22:50
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hint
$$p-1\equiv -1 \mod p$$
So, the remainder after a division by $ p $, won't be $ -1 $ but $ p-1$.
hamam_Abdallah
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I understood that p-1 $\equiv$ -1(modp)..But I didn't understand the next step – Gitika Aug 09 '20 at 15:23
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@Gitika The remainder is non negative. it could not be $-1$ but $ p-1$. Take an example. – hamam_Abdallah Aug 09 '20 at 22:19