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  "body": "\n## Statement\n\nLet $n$ be a positive integer, let $a$ and $b$ be integers, and let $k$ be a natural number.\nIf $a \\equiv b \\pmod n$, then $a^k \\equiv b^k \\pmod n$.\n\n## Remarks\n\nThe proof is an induction on $k$ with [[compatibility with multiplication|072D5G4SK6]].\nThe exponent cannot be reduced modulo $n$ in general. For example, $2^5 = 32 \\equiv 2 \\pmod 3$, but $2^{5 \\bmod 3} = 2^2 \\equiv 1 \\pmod 3$.\n[[Euler's theorem|0VC4DVE7DQ]] tells when one can reduce the exponent, and by which modulus.\n\n## Explanations\n\n### L1 human\n\nTo find $10^{100} \\bmod 9$, note that $10 \\equiv 1 \\pmod 9$.\nSo $10^{100} \\equiv 1^{100} = 1 \\pmod 9$, without a computation of the large number.\n",
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