theorem
  for N being sequence of NAT holds (s * N) ^\k = s * (N ^\k)
proof
  let N be sequence of NAT;
  now
    let n be Element of NAT;
    thus ((s * N) ^\k).n = (s * N).(n + k) by Def2
      .= s.(N.(n + k)) by FUNCT_2:15,ORDINAL1:def 12
      .= s.((N ^\k).n) by Def2
      .= (s * (N ^\k)).n by FUNCT_2:15;
  end;
  hence thesis by FUNCT_2:63;
end;
