reserve a, b, n for Nat,
  r for Real,
  f for FinSequence of REAL;
reserve p for Prime;

theorem Th25:
  1 < b implies b |-count (b|^a) = a
proof
  reconsider a as Element of NAT by ORDINAL1:def 12;
  assume
A1: b > 1;
  then reconsider b as non zero Element of NAT by ORDINAL1:def 12;
  now
    b|^a divides b|^(a+1) by NAT_1:11,NEWTON:89;
    then
A2: b|^a <= b|^(a+1) by NAT_D:7;
    assume b|^(a+1) divides b|^a;
    then b|^(a+1) <= b|^a by NAT_D:7;
    then b|^a = b|^(a+1) by A2,XXREAL_0:1;
    then a+0 = a+1 by A1,PEPIN:30;
    hence contradiction;
  end;
  hence thesis by A1,Def7;
end;
