Remainders in Pascal's triangle modulo a prime p: Suppose n = ∑nᵢpⁱ and k = ∑kᵢpⁱ. In 𝔽ₚ[x],
∑ₖ (nCk)xᵏ = (1+x)ⁿ = ∏ᵢ (1+x)^{nᵢpⁱ} = ∏ᵢ (1+x^{pⁱ})^{nᵢ}
with final equality by Frobenius. Expanding the RHS and equating coeff's of xᵏ gives
(nCk) ≡ ∏ᵢ (nᵢCkᵢ) (mod p).
