
theorem
  for p be Prime, n be Nat holds
    ((p + n) choose p) mod p = ((n div p) + 1) mod p
  proof
    let p be Prime, n be Nat;
    (1*p + n) mod p = n mod p & (1*p + 0) mod p = 0 mod p &
    (1*p + n) div p = (n div p) + 1 &
      (1*p + 0) div p = (0 div p) + 1 by NAT_D:61; then
    ((p + n) choose p) mod p
    = (((n mod p) choose 0)*((1 + (n div p)) choose 1)) mod p by AL
    .= (1*((n div p) + 1)) mod p by NEWTON:19;
    hence thesis;
  end;
