reserve d,i,j,k,m,n,p,q,x,k1,k2 for Nat,
  a,c,i1,i2,i3,i5 for Integer;

theorem Th33:
  for n st
  p is prime & n < p |^ (k+1) holds n divides p |^ (k+1) iff n divides p |^ k
proof
  let n;
  assume that
A1: p is prime and
A2: n < p |^ (k+1);
A3: p <> 0 by A1,INT_2:def 4;
A4: n divides p |^ (k+1) implies n divides p |^ k
  proof
    assume
A5: n divides p |^ (k+1);
    now
      per cases by A1,A5,Th32;
      suppose
        n = 1;
        hence thesis by NAT_D:6;
      end;
      suppose
        ex i being Element of NAT st n = p*i;
        then consider i being Element of NAT such that
A6:     n = p*i;
        consider u being Nat such that
A7:     p |^ (k+1) = (p*i)*u by A5,A6,NAT_D:def 3;
        p*(p |^ k) = p*(i*u) by A7,NEWTON:6;
        then
A8:     p |^ k = p*(i*u) div p by A3,NAT_D:18;
        then
A9:     p |^ k = i*u by A3,NAT_D:18;
        then
A10:    u divides p |^ k;
        u <> 1 by A2,A6,A9,NEWTON:6;
        then consider j being Element of NAT such that
A11:    u = p*j by A1,A10,Th32;
        p |^ k = n*j by A3,A6,A8,A11,NAT_D:18;
        hence thesis;
      end;
    end;
    hence thesis;
  end;
  n divides p |^ k implies n divides p |^ (k+1)
  proof
    assume n divides p |^ k;
    then n divides (p |^ k)*p by NAT_D:9;
    hence thesis by NEWTON:6;
  end;
  hence thesis by A4;
end;
