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  "body": "\n## Proof\n\n1. Let $p$ be a [[prime number|0SN0MAEKYX]] and let $k$ be a positive integer. By definition, [[Euler's totient function|0VV04M41AN]] $\\varphi(p^k)$ is the number of integers $x$ with $1 \\le x \\le p^k$ that are [[coprime|0GFKP8F8AG]] to $p^k$.\n2. We show that an integer $x$ is coprime to $p^k$ if and only if $p$ does not [[divide|096PW7Q014]] $x$.\n3. Assume $p \\mid x$. Since $k \\ge 1$, $p$ also divides $p^k$. So $p$ is a common divisor, and $\\gcd(x, p^k) \\ge p \u003e 1$.\n4. Assume $p \\nmid x$, and let $d = \\gcd(x, p^k)$ be the [[greatest common divisor|02WF878BH1]]. Since $p^k \\ne 0$, $d$ is positive. Suppose that $d \u003e 1$.\n5. By the [[existence of a prime divisor|0SPBEDVNW1]], there is a prime $q$ with $q \\mid d$. Then $q$ divides $p^k = p \\cdot p \\cdots p$.\n6. By [[Euclid's lemma|051VYKA1EX]], applied to the $k$ factors in turn, $q$ divides $p$. Since $q \u003e 1$ and $p$ is prime, $q = p$.\n7. Then $p = q$ divides $d$, and $d$ divides $x$, so $p \\mid x$. This contradicts step 4. So $d = 1$.\n8. By steps 3 and 7, the integers in $\\{ 1, \\dots, p^k \\}$ that are not coprime to $p^k$ are exactly the multiples of $p$ in this set.\n9. These multiples are $p j$ with $1 \\le j \\le p^{k-1}$, since $1 \\le p j \\le p^k$ is equivalent to $1 \\le j \\le p^{k-1}$ for an integer $j$. So there are $p^{k-1}$ of them.\n10. Hence $\\varphi(p^k) = p^k - p^{k-1} = p^{k-1} (p - 1)$.\n\n## Idea\n\nThe only prime factor of $p^k$ is $p$, so a number shares a factor with $p^k$ exactly when $p$ divides it. Among $1, \\dots, p^k$, every $p$-th number is a multiple of $p$, so we remove $p^{k-1}$ of them.\n",
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