theorem Th78:
  n is having_exactly_one_prime_divisor implies
  ex k being non zero Nat st n = (the_only_divisor_of n) |^ k
  proof
    assume
A1: n is having_exactly_one_prime_divisor;
    then consider p being Prime such that
A2: p divides n and
A3: for r being Prime st r <> p holds not r divides n;
A4: the_only_divisor_of n = p by A1,A2,Def7;
    now
      assume for k being Nat holds n <> p|^k;
      then ex s being Element of NAT st
      s is prime & s divides n & s <> p by GROUPP_1:1;
      hence contradiction by A3;
    end;
    then consider k being Nat such that
A5: n = (the_only_divisor_of n) |^ k by A4;
    reconsider k as non zero Nat by A1,A5;
    take k;
    thus thesis by A5;
  end;
