theorem Th2:
  (s1^\k)(#)(seq^\k)= (s1(#)seq) ^\k
  proof
    now let n;
      thus ((s1(#)seq) ^\k).n=(s1(#)seq).(n+k) by NAT_1:def 3
      .=s1.(n+k)*seq.(n+k) by NDIFF_1:def 2
      .=(s1 ^\k).n*seq.(n+k) by NAT_1:def 3
      .=(s1 ^\k).n*(seq ^\k).n by NAT_1:def 3
      .=((s1 ^\k)(#)(seq ^\k)).n by NDIFF_1:def 2;
    end;
    hence thesis by FUNCT_2:63;
  end;
