
theorem DivK:
  for k being Nat, p being Prime st k >= 1 holds
    PrimeDivisors (p |^ k) = {p}
  proof
    let k be Nat, p be Prime;
    assume
A0: k >= 1;
    defpred P[Nat] means PrimeDivisors (p |^ ($1+1)) = {p};
A1: P[0] by LemmaOne;
A2: for k being Nat st P[k] holds P[k+1]
    proof
      let k be Nat;
      assume
A3:   P[k];
      PrimeDivisors (p |^ (k+1+1)) = PrimeDivisors (p |^ (k+1) * p)
                    by NEWTON:6
             .= PrimeDivisors (p |^ (k+1)) \/ PrimeDivisors p by DivisorsMulti
             .= PrimeDivisors (p |^ (k+1)) \/ {p} by LemmaOne
             .= {p} by A3;
      hence thesis;
    end;
    for n being Nat holds P[n] from NAT_1:sch 2(A1,A2); then
    P[k-'1];
    hence thesis by A0,XREAL_1:235;
  end;
