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  "body": "\n## Proof\n\n1. Let $p$ be a [[prime number|0SN0MAEKYX]]. By definition, [[Euler's totient function|0VV04M41AN]] $\\varphi(p)$ is the number of integers $x$ with $1 \\le x \\le p$ that are [[coprime|0GFKP8F8AG]] to $p$.\n2. Let $x$ be an integer with $1 \\le x \\le p$, and let $d = \\gcd(x, p)$ be the [[greatest common divisor|02WF878BH1]]. It is a positive divisor of $p$.\n3. Since $p$ is prime, its only positive divisors are $1$ and $p$. So $d = 1$ or $d = p$.\n4. If $x \\le p - 1$, then $0 \u003c x \u003c p$, so $p$ does not divide $x$. Since $d$ divides $x$, $d \\ne p$, and so $d = 1$.\n5. If $x = p$, then $d = \\gcd(p, p) = p$. Since $p \u003e 1$, $p$ is not coprime to itself.\n6. By steps 4 and 5, the integers in $\\{ 1, \\dots, p \\}$ that are coprime to $p$ are exactly $1, 2, \\dots, p - 1$.\n7. Hence $\\varphi(p) = p - 1$.\n\n## Idea\n\nA prime has no divisors other than $1$ and itself. So every positive integer below $p$ shares no factor with $p$, and only $p$ itself is left out.\n",
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