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Show that if $p$ is a odd prime number, then $p\mid 2^{p-1} + (p -1)!$

Had this problem on a list of exercíses but I don't know what to do, I would guess the use of Wilson's theorem however I am not accustomed enought to its aplication for the solution, so if anyone could explain it I would appreciate it.

Bernard
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BittenBrittle
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    $1-1=0{}{}{}{}{}$ – markvs Nov 21 '21 at 18:00
  • As @markvs hinted, you also need Fermat's Little Theorem. That's why the problem is for odd primes. – J.G. Nov 21 '21 at 18:14
  • Similar to the linked dupe, apply the little Fermat and Wilson Theorems to the summands, reducing the sum to $,1 + (-1)\equiv 0\pmod{p},$ via the Congruence Sum Rule. Equivalently: note $2^{p-1}-1$ and $(p-1)!+1$ are both divisible by $p$ hence so too is their sum. – Bill Dubuque Nov 21 '21 at 18:18
  • So I read the duplicate, and please correct me if I am wrong: By applying FLT we have that 2^p-1 ≡ 1 *mod p) and Wilson's theorem says that (p−1)!≡−1(modp), so we would have p| 1 (mod p) - 1 (mod p) <=> 1 (mod p) - 1 (mod p) (mod p), but where do I gor from here? – BittenBrittle Nov 21 '21 at 18:29
  • By the congruence sum rule the sum is $\equiv 1+(-1)\equiv 0\pmod{p},$ which means it is divisible by $p\ \ $ – Bill Dubuque Nov 21 '21 at 18:41
  • I think in exsence what I mean to ask that just by proving that 1 + (-1) ≡ 0 (mod p) is the solution? or does it need anything else? I am sorry for the question but I am new to the subject so I just want to be sure, just saw your awnser @Bill Dubuque, so would that explanation sufice as the conclusion? – BittenBrittle Nov 21 '21 at 18:41
  • Forget I asked now that I used my brain a bit more it does, thank you everyone for the help – BittenBrittle Nov 21 '21 at 18:47

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