
theorem for p be Prime, n be Nat holds ((p*n) choose p) mod p = n mod p
  proof
    let p be Prime, n be Nat;
    per cases;
    suppose n is zero;
      hence thesis;
    end;
    suppose n is non zero; then
      reconsider n as non zero Nat;
      (n*p + 0) mod p = 0 mod p & (1*p + 0) mod p = 0 mod p &
      (n*p + 0) div p = (0 div p) + n &
        (1*p + 0) div p = (0 div p) + 1 by NAT_D:61; then
      ((p*n) choose p) mod p = ((0 choose 0)*(n choose 1)) mod p by AL
      .= (1*n) mod p by NEWTON:21;
      hence thesis;
    end;
  end;
