theorem
  (z * seq) ^\k = z * (seq ^\k)
proof
  now
    let n be Element of NAT;
    thus ((z * seq) ^\k).n = (z * seq).(n + k) by NAT_1:def 3
      .= z * (seq.(n + k)) by CLVECT_1:def 14
      .= z * ((seq ^\k).n) by NAT_1:def 3
      .= (z * (seq ^\k)).n by CLVECT_1:def 14;
  end;
  hence thesis by FUNCT_2:63;
end;
