My book says that the statement "$\forall a \not \equiv 0 (\mod p)$ and $p$ prime, $a^{p-1} \equiv 1 (\mod p)$" follows from Wilson's Theorem. I'd like to know how. This is what I've looked at so far:

Since $p$ is prime, $(p-1)! \equiv -1 (\mod p) \Rightarrow a^{(p-1)!} \equiv a^{-1} (\mod p) \Rightarrow a^{(p-1)!+1}\equiv 1(\mod p)$. I cannot see how to arrive at the statement from here.

  • 2
    $\begingroup$ Look at $$\prod_{k=1}^{p-1} (a\cdot k).$$ Besides, you can't replace exponents with their remainders modulo $p$. After you have Fermat's theorem, you can replace them with their remainders modulo $p-1$. $\endgroup$ – Daniel Fischer Jul 31 '14 at 18:01

Observe that $S:\{b_i\}, T:\{a\cdot b_i\} 1\le i\le p-1$ are the sets of reduced residues if $(a,p)=1$

$$\implies\prod_{i=1 }^{p-1} b_i\equiv \prod_{i=1 }^{p-1} a\cdot b_i\pmod p$$


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