theorem Th21:
  (seq(#)seq1)^\k = (seq^\k)(#)(seq1^\k)
proof
  now
    let n be Element of NAT;
    thus ((seq(#)seq1)^\k).n = (seq(#)seq1).(n+k) by NAT_1:def 3
      .= seq.(n+k)*seq1.(n+k) by VALUED_1:5
      .= (seq^\k).n*seq1.(n+k) by NAT_1:def 3
      .= (seq^\k).n*(seq1^\k).n by NAT_1:def 3
      .= ((seq^\k)(#)(seq1^\k)).n by VALUED_1:5;
  end;
  hence thesis;
end;
