theorem
  (a * seq) ^\k = a * (seq ^\k)
proof
  now
    let n be Element of NAT;
    thus ((a * seq) ^\k).n = (a * seq).(n + k) by NAT_1:def 3
      .= a * (seq.(n + k)) by NORMSP_1:def 5
      .= a * ((seq ^\k).n) by NAT_1:def 3
      .= (a * (seq ^\k)).n by NORMSP_1:def 5;
  end;
  hence thesis by FUNCT_2:63;
end;
