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  "body": "\n## Proof\n\nWe argue by induction on $k$.\n\n1. Let $n$ be a positive integer, and let $a$ and $b$ be integers with $a \\equiv b \\pmod n$.\n2. Base case $k = 0$: we have $a^0 = 1 = b^0$. Since $n$ [[divides|096PW7Q014]] $1 - 1 = 0$, we get $a^0 \\equiv b^0 \\pmod n$ by the definition of [[congruence|0TPW25CBN6]].\n3. Induction step: let $k \\ge 0$ and assume $a^k \\equiv b^k \\pmod n$.\n4. By step 3, by step 1 and by [[compatibility with multiplication|072D5G4SK6]], $a^k \\cdot a \\equiv b^k \\cdot b \\pmod n$.\n5. That is, $a^{k+1} \\equiv b^{k+1} \\pmod n$.\n6. By steps 2 to 5 and induction, $a^k \\equiv b^k \\pmod n$ for each natural number $k$.\n\n## Idea\n\nA power is a repeated product. Each new factor $a$ can be replaced by $b$ without changing the class modulo $n$, by compatibility with multiplication, so induction on the exponent finishes the proof.\n",
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