theorem Th59:
  for k be Nat st k+1 is prime & not k+1 divides a
    holds k+1 divides a|^k - 1
  proof
    let k be Nat such that
    A1: k+1 is prime and
    A2: not k+1 divides a;
    k+1 divides a|^(k+1) - a by A1,Th58; then
    k+1 divides a|^k*a - a by NEWTON:6; then
    k+1 divides a*(a|^k - 1);
    hence thesis by A1,A2,INT_5:7;
  end;
