
theorem lemdiv:
  for a being Integer, m being Nat holds a-1 divides a|^m - 1
  proof
    let a be Integer, m be Nat;
    defpred P[Nat] means a-1 divides a|^$1 - 1;
    a|^0 - 1 = 1-1 by NEWTON:4;
    then z: P[0] by INT_2:12;
    i:
    now
      let k be Nat;
      assume P[k];
      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 P[k+1] by INT_1:def 3;
    end;
    for k being Nat holds P[k] from NAT_1:sch 2(z,i);
    hence thesis;
  end;
