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Why is $g^e \mod p = g^{e \mod (p-1)} \mod p$ if p is prime. I don't get it. It follows from Fermat's little theorem.

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Fermat's little theorem $$a^p \equiv a \pmod{p} \Leftrightarrow a^{p-1} \equiv 1 \pmod{p}(p \nmid a)$$

so next divides $e$ with $p-1$:$$e=(p-1)r+s;$$ $$s\equiv e \pmod{p-1};$$ so we get $$g^{(p-1)r+s} \pmod {p} \equiv (g^{p-1})^rg^s \pmod{p}$$ follow Fermat's little theorem $g^{p-1} \equiv 1\pmod{p}$ finally we get :$$g^{(p-1)r+s} \pmod {p} \equiv (g^{p-1})^rg^s \pmod{p} \equiv g^s \pmod{p}$$ $\Leftrightarrow$ $$g^e \pmod{p} \equiv g^{e \pmod{p-1}} \pmod{p}$$

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Thanks Alex!!!! – user4811 Sep 10 '13 at 1:34
@Alex For the left-to-right implication in the first equivalence to hold true, you should probably add that $p \nmid a$. – hakoja Sep 10 '13 at 9:32
thanks yes you are right – T.B Sep 10 '13 at 10:10

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