
theorem
  for p be odd Prime, k be non zero Nat st k + 1 < p holds
    p divides ((p + 1) choose (k + 1))
  proof
    let p be odd Prime, k be non zero Nat such that
    A1: k + 1 < p;
    0 < k + 0 < k + 1 < p by A1,XREAL_1:6; then
    p divides (p choose k) & p divides (p choose (k + 1)) by NEWTON02:119; then
    p divides ((p choose k) + (p choose (k + 1))) by WSIERP_1:4;
    hence thesis by NEWTON:22;
  end;
