September 17th
Today I learned a very natural proof of Lucas's Theorem: for $p$ prime and integers $m=\sum_km_kp^k$ and $n=\sum_kn_kp^k$ written in base $p,$ we have\[\binom mn\equiv\prod_{k=0}^\infty\binom{m_k}{n_k}\pmod p.\]
What I really like about this proof is that it's really just an answer to "What happens if I apply the Frobenius to base-$p$ expansions?'' Everything else follows naturally. Indeed, the main ingredient in the proof is generating functions. Observe that\[(1+x)^p=1+x^p\pmod p\]in $\FF_p[x].$ From this it follwos $(1+x)^{p^k}=1+x^{p^k}$ by induction. So to prove the result, we combine base-$p$ representations with this automorphism. On one hand, we see\[(1+x)^m=\sum_{n=0}^m\binom mnx^n.\]It follows in the resulting discussion we know that we want to isolate $x^n.$ So on the other hand, we can write\[(1+x)^m=(1+x)^{\sum_km_kp^k}=\prod_{k=0}^\infty(1+x)^{m_kp^k}.\]But applying the Frobenius automorphism, this looks like\[(1+x)^m=\prod_{k=0}^\infty\left(1+x^{p^k}\right)^{m_k}=\prod_{k=0}^\infty\left(\sum_{n_k=0}^{m_k}\binom{m_k}{n_k}x^{n_kp^k}\right)\]after expanding. Now because $m_k \lt p,$ we can extend this into\[(1+x)^m=\prod_{k=0}^\infty\left(\sum_{n_k=0}^{p-1}\binom{m_k}{n_k}x^{n_kp^k}\right).\]Now we can collect powers of $x$ easily because $x^n$ will have a unique base-$p$ expansion from the $n_k,$ giving\[(1+x)^m=\sum_{n=0}^\infty\left(\prod_{k=0}^\infty\binom{m_k}{n_k}\right)x^n.\]Comparing coefficients with our original expression gives the result.