let a be Integer; :: thesis: for m being Nat holds a - 1 divides (a |^ m) - 1
let m be Nat; :: thesis: a - 1 divides (a |^ m) - 1
defpred S₁[ Nat] means a - 1 divides (a |^ $1) - 1;
(a |^ 0) - 1 = 1 - 1 by NEWTON:4;
then z: S₁[ 0 ] by INT_2:12;

i: now :: thesis: for k being Nat st S₁[k] holds
S₁[k + 1]

let k be Nat; :: thesis: ( S₁[k] implies S₁[k + 1] )
assume S₁[k] ; :: thesis: S₁[k + 1]
then consider l being Integer such that
l: (a |^ k) - 1 = (a - 1) * l by INT_1:def 3;
(a |^ (k + 1)) - 1 = ((a |^ k) * a) - 1 by NEWTON:6
.= (a - 1) * ((a |^ k) + l) by l ;
hence S₁[k + 1] by INT_1:def 3; :: thesis: verum

end;

for k being Nat holds S₁[k] from NAT_1:sch 2(z, i);
hence a - 1 divides (a |^ m) - 1 ; :: thesis: verum