1 and p i s prime, then P ^ g, . \\ 16. Proof ; g k = g.c.d{(J)} 1=1,2 (k-1) . To show p ^ g^ we need only f i n d one i such that p \\ (^). \u2022 cc Consider the case i n which i = p . P a x Let us evaluate N (( )) P P Since ( ? c c x ) - (p g x! ) V P \u00ab ! (( x - l ) p \u00ab ) ! ' . C we have N (