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If $p$ is a prime, then $(p-1)! \equiv -1 (mod p)$. Hint: $(p-1)!$ is the product of elements in $Z_p$. Match each element to its inverse.

I can understand by testing some primes that for any prime $(p-1)!+1$ is a multiple of that prime, but I'm not sure how to prove it in general or how inverses play into it.

d.v.
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    What about the hint given to you? "How do inverses play into it" well, inverses cancel each other out, so you need to locate the inverses in the product $(p-1)!$, and see what cancels out, and what does not. Some elements in that product have an inverse in that product. Some, do not. – Sarvesh Ravichandran Iyer Oct 02 '20 at 03:29
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    Look up Wilson's Theorem either via google or in any intro number theory book. [that's the displayed statement] – coffeemath Oct 02 '20 at 03:29
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    There are only even number amount of elements in the group, and $a=a^{-1}$ iff $a=\pm1$. – Lynnx Oct 02 '20 at 03:31
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    It is Wilson's Theorem and the converse is also true –  Oct 02 '20 at 03:31
  • I'm assuming this is a group theory course, and OP might not want a number theory proof. – Lynnx Oct 02 '20 at 03:32
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    i made it too complicated for myself. thank you everyone – d.v. Oct 02 '20 at 03:50

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