theorem Th36:
  Sigma(k) is multiplicative
proof
  for n being non zero Nat holds (Sigma k).n = Sum((EXP k)|
  NatDivisors n)
  proof
    let n be non zero Nat;
    thus (Sigma k).n = sigma(k,n) by Def3
      .= Sum((EXP k)|NatDivisors n) by Def2;
  end;
  hence Sigma(k) is multiplicative by Th34,Th35;
end;
