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  "body": "\n## Proof\n\nWe count the elements of each order and compare with the totient function.\n\n1. Let $p$ be a prime and let $G = (\\Z/p\\Z)^\\times$ be the [[group of units|0F8X8ZRQ53]] of $\\Z/p\\Z$. Write $1$ for its identity $[1]$.\n2. The ring $\\Z/p\\Z$ is a [[field|0NV6E43P4A]], so each nonzero class is a [[unit|019NFGFPR5]].\n3. The class $[0]$ is not a unit, because $[0][x] = [0] \\neq [1]$ for each $x$, as $p \\geq 2$. So $G$ is the set of the $p - 1$ nonzero classes, and $\\abs{G} = p - 1$.\n4. For each positive divisor $d$ of $p - 1$, let $N(d)$ be the number of elements of $G$ of [[order|0WE2NHZJ7B]] $d$.\n5. Each element of the finite group $G$ has an order, and this order [[divides|08V7JHS3QZ]] $\\abs{G} = p - 1$. So $\\sum_{d \\mid p - 1} N(d) = p - 1$.\n6. Fix a positive divisor $d$ of $p - 1$. We show that $N(d) \\leq \\varphi(d)$, where $\\varphi$ is [[Euler's totient function|0VV04M41AN]]. If $N(d) = 0$, this holds, since $\\varphi(d) \\geq 0$.\n7. Otherwise, let $a \\in G$ have order $d$. The powers $a^0, a^1, \\ldots, a^{d-1}$ are $d$ [[distinct elements|031M8HX7DA]] of $G$.\n8. Each power satisfies $(a^k)^d = (a^d)^k = 1$. So each of them is a root of the polynomial $T^d - 1$ in $\\Z/p\\Z$.\n9. The polynomial $T^d - 1$ is nonzero of [[degree|0DPF6WRZW6]] $d$. Since $\\Z/p\\Z$ is a field, it has [[at most $d$ roots|0KJVQV3E0F]] in $\\Z/p\\Z$.\n10. By steps 7 to 9, the roots of $T^d - 1$ in $\\Z/p\\Z$ are exactly $a^0, \\ldots, a^{d-1}$.\n11. Let $b \\in G$ have order $d$. Then $b^d = 1$, so $b$ is a root of $T^d - 1$. By step 10, $b = a^k$ for some $k$ with $0 \\leq k \\leq d - 1$.\n12. By the [[order of a power|0VVPQNVM8S]], $a^k$ has order $d / \\gcd(d, k)$. This equals $d$ if and only if $\\gcd(d, k) = 1$.\n13. The powers in step 7 are distinct, so $N(d)$ is the number of $k$ with $0 \\leq k \\leq d - 1$ and $\\gcd(d, k) = 1$.\n14. Since $\\gcd(d, 0) = d = \\gcd(d, d)$, we may replace $k = 0$ by $k = d$. So $N(d)$ is the number of $k$ with $1 \\leq k \\leq d$ and $\\gcd(d, k) = 1$, which is $\\varphi(d)$.\n15. Thus $N(d) \\leq \\varphi(d)$ for every positive divisor $d$ of $p - 1$.\n16. By step 5, step 15 and the [[sum of the totient over the divisors|09JXE6XMCC]], $p - 1 = \\sum_{d \\mid p - 1} N(d) \\leq \\sum_{d \\mid p - 1} \\varphi(d) = p - 1$.\n17. So the inequality in step 16 is an equality. Each term satisfies $N(d) \\leq \\varphi(d)$, so $N(d) = \\varphi(d)$ for each divisor $d$ of $p - 1$.\n18. In particular, $N(p - 1) = \\varphi(p - 1)$. This is at least $1$, because $k = 1$ satisfies $1 \\leq 1 \\leq p - 1$ and $\\gcd(p - 1, 1) = 1$.\n19. So there is an element $g \\in G$ of order $p - 1$. The subgroup $\\langle g \\rangle$ has $p - 1$ [[distinct elements|031M8HX7DA]].\n20. Since $\\langle g \\rangle \\subseteq G$ and both sets have $p - 1$ elements, $\\langle g \\rangle = G$.\n21. Hence $G = (\\Z/p\\Z)^\\times$ is [[cyclic|0KVGH0J5B3]] of order $p - 1$.\n\n## Idea\n\nIn a field, the equation $x^d = 1$ has at most $d$ solutions. So if one element of order $d$ exists, all elements of order $d$ lie among its powers, and there are exactly $\\varphi(d)$ of them. Thus there are at most $\\varphi(d)$ elements of order $d$. Since the totients of the divisors of $p - 1$ add up to $p - 1$, no order can be missing, not even $p - 1$.\n",
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