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  "body": "\n## Proof\n\nWe split the [[Cartesian product|0C81YBVTWH]] into disjoint copies of $A$, and then argue by induction on $k$.\n\n1. Let $m = |A|$ and $n = |B|$. By the definition of [[cardinality|05099E1NMJ]], there are [[bijections|0SNQEF7Z53]] $f : \\set{1, \\dots, m} \\to A$ and $g : \\set{1, \\dots, n} \\to B$. Write $b_j = g(j)$.\n2. If $n = 0$, then $B$ is empty, so $A \\times B$ is empty and $|A \\times B| = 0 = m \\cdot 0$.\n3. Let $n \\ge 1$. Every pair $(a, b) \\in A \\times B$ has $b = b_j$ for exactly one $j$. So $A \\times B$ is the union of the sets $A \\times \\set{b_j}$ for $j = 1, \\dots, n$.\n4. These sets are pairwise disjoint, because $b_j \\ne b_{j'}$ for $j \\ne j'$, as $g$ is injective.\n5. For each $j$, the map $i \\mapsto (f(i), b_j)$ from $\\set{1, \\dots, m}$ to $A \\times \\set{b_j}$ is a bijection, because $f$ is a bijection. So $|A \\times \\set{b_j}| = m$.\n6. By the [[rule of sum|0EE2V3CHFK]], $A \\times B$ is finite and $|A \\times B| = m + \\dots + m = n m = |A| \\cdot |B|$, with $n$ terms.\n7. Now let $A_1, \\dots, A_k$ be finite sets, with $k \\ge 1$. For $k = 1$ the claim $|A_1| = |A_1|$ holds.\n8. Suppose the claim holds for $k$ sets. Put $P = A_1 \\times \\dots \\times A_k$. By hypothesis $P$ is finite and $|P| = |A_1| \\cdots |A_k|$.\n9. The map $\\phi : P \\times A_{k+1} \\to A_1 \\times \\dots \\times A_{k+1}$ with $\\phi((a_1, \\dots, a_k), a_{k+1}) = (a_1, \\dots, a_{k+1})$ is a bijection. Its inverse regroups the entries.\n10. By steps 1 to 6, there is a bijection $h : \\set{1, \\dots, N} \\to P \\times A_{k+1}$ with $N = |P| \\cdot |A_{k+1}|$.\n11. A composite of bijections is a bijection, since [[injections compose|01FPXMTBX1]] and [[surjections compose|07EH4R3RVQ]]. So $\\phi \\circ h$ is a bijection from $\\set{1, \\dots, N}$ to $A_1 \\times \\dots \\times A_{k+1}$.\n12. Hence $|A_1 \\times \\dots \\times A_{k+1}| = N = |A_1| \\cdots |A_k| \\cdot |A_{k+1}|$. This completes the induction.\n\n## Idea\n\nThe pairs with a fixed second entry $b$ form a copy of $A$.\nThere are $|B|$ such copies and they do not overlap, so the rule of sum adds $|A|$ to itself $|B|$ times.\nLonger products are built one factor at a time.\n",
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