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  "body": "\n## Proof\n\n1. Let $p$ be a [[prime number|0SN0MAEKYX]] and $m = \\sum_{i=0}^{k} m_i p^i$, $n = \\sum_{i=0}^{k} n_i p^i$ with digits $0 \\le m_i, n_i \u003c p$.\n2. Let $R = (\\mathbb{Z}/p\\mathbb{Z})[x]$ be the [[polynomial ring|0Q9EZS90GY]] over the [[integers modulo p|01N3AYAA6N]]. It is a commutative ring. In $R$, the sum of $j$ copies of $1$ is the class of $j$, which is $0$ for $j = p$ and nonzero for $0 \u003c j \u003c p$.\n3. So $R$ has [[characteristic|0P7CH1CND5]] $p$. By the [[freshman's dream|0396VGFA78]], $(1 + x)^p = 1 + x^p$ in $R$.\n4. By induction on $i$, $(1 + x)^{p^i} = 1 + x^{p^i}$ in $R$. The case $i = 0$ is clear. If it holds for $i$, then $(1+x)^{p^{i+1}} = (1 + x^{p^i})^p = 1 + x^{p^{i+1}}$, by the freshman's dream applied to $1$ and $x^{p^i}$.\n5. By the [[binomial theorem|0EHWPWGHB2]] in $R$, for each natural number $N$, $(1 + x)^N = \\sum_{j=0}^{N} \\overline{\\binom{N}{j}} x^j$. Here the bar is the class modulo $p$ of the [[binomial coefficient|092GEF38YD]].\n6. So the coefficient of $x^n$ in $(1+x)^m$ is the class of $\\binom{m}{n}$. This also holds when $n \u003e m$, since the coefficient is then $0$ and $\\binom{m}{n} = 0$ by convention.\n7. Since $m = \\sum_i m_i p^i$, step 4 gives $(1 + x)^m = \\prod_{i=0}^{k} \\left((1+x)^{p^i}\\right)^{m_i} = \\prod_{i=0}^{k} (1 + x^{p^i})^{m_i}$.\n8. By the binomial theorem, $(1 + x^{p^i})^{m_i} = \\sum_{j=0}^{m_i} \\overline{\\binom{m_i}{j}} x^{j p^i}$.\n9. Multiplying out, $(1 + x)^m$ is the sum, over all tuples $(j_0, \\dots, j_k)$ with $0 \\le j_i \\le m_i$, of $\\overline{\\prod_i \\binom{m_i}{j_i}} \\, x^{\\sum_i j_i p^i}$.\n10. Base-$p$ digits are unique. If $\\sum_i a_i p^i = \\sum_i b_i p^i$ with digits in $\\set{0, \\dots, p-1}$, the [[division algorithm|0CHQKZFK5S]] by $p$ gives $a_0 = b_0$. Subtracting, dividing by $p$ and repeating gives $a_i = b_i$ for all $i$.\n11. In step 9 each $j_i \\le m_i \u003c p$. So by step 10, $\\sum_i j_i p^i = n$ holds exactly when $j_i = n_i$ for every $i$.\n12. If $n_i \\le m_i$ for all $i$, exactly one tuple contributes to $x^n$, and the coefficient of $x^n$ is the class of $\\prod_i \\binom{m_i}{n_i}$.\n13. If $n_i \u003e m_i$ for some $i$, no tuple contributes, and the coefficient of $x^n$ is $0$. In this case $\\binom{m_i}{n_i} = 0$, so the coefficient is again the class of $\\prod_i \\binom{m_i}{n_i}$.\n14. Two equal polynomials have equal coefficients. Comparing steps 6, 12 and 13 gives $\\binom{m}{n} \\equiv \\prod_{i=0}^{k} \\binom{m_i}{n_i} \\pmod p$, a [[congruence modulo p|0TPW25CBN6]].\n\n## Idea\n\nModulo $p$, raising to the power $p$ is additive, so $(1+x)^{p^i} \\equiv 1 + x^{p^i}$. Writing $m$ in base $p$ factors $(1+x)^m$ into pieces, one for each digit. Uniqueness of base-$p$ digits means that $x^n$ arises from exactly one choice in each piece, namely the digit $n_i$.\n",
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