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  "body": "\n## Proof\n\n1. Call a finite chain of subfields $\\Q = F_0 \\subseteq F_1 \\subseteq \\dots \\subseteq F_m$ of $\\C$ a square-root tower if, for each $j$, [[$F_j = F_{j-1}(s_j)$|0MA5B744K3]] for some $s_j \\in \\C$ with $s_j^2 \\in F_{j-1}$. The case $m = 0$ is allowed.\n2. Let $T$ be the set of the complex numbers that lie in the top field $F_m$ of some square-root tower.\n3. In a square-root tower, $s_j$ is a root of $x^2 - s_j^2 \\in F_{j-1}[x]$. So its [[minimal polynomial|0GK1WB9R98]] over $F_{j-1}$ [[divides|06K7SARZKR]] this polynomial and has degree $1$ or $2$.\n4. By the [[degree of a simple extension|0N5QW9K7MH]], $[F_j : F_{j-1}] \\in \\set{1, 2}$. By the [[tower law|0A7KJ68XHT]], $[F_m : \\Q]$ is a product of $1$s and $2$s, so it is a power of $2$.\n5. Let $z, w \\in T$, with towers $(F_j)_{j \\le m}$ built from $s_1, \\dots, s_m$ and $(G_j)_{j \\le l}$ built from $t_1, \\dots, t_l$.\n6. Extend the first tower by $F_{m+j} = F_{m+j-1}(t_j)$ for $1 \\le j \\le l$. By induction on $j$, $G_j \\subseteq F_{m+j}$: indeed $G_j = G_{j-1}(t_j)$, and $F_{m+j}$ contains $G_{j-1}$ and $t_j$.\n7. So $t_j^2 \\in G_{j-1} \\subseteq F_{m+j-1}$, and the extended chain is a square-root tower. Its top field contains $z$ and $w$.\n8. That top field is a field. So $z + w$, $-z$, $zw$, and $z^{-1}$ for $z \\neq 0$, lie in it and hence in $T$. Also $\\Q \\subseteq T$, from the tower with $m = 0$. So $T$ is a [[subfield|07QY5RPFBR]] of $\\C$.\n9. Let $z \\in T$ lie in the top field $F_m$ of a tower, and let $s \\in \\C$ with $s^2 = z$. Then $F_{m+1} = F_m(s)$ extends the tower, so $s \\in T$. So $T$ is closed under square roots.\n10. The [[constructible numbers|0FMN2MN7TM]] form the [[smallest subfield of $\\C$ closed under square roots|0PN9WMRWT7]]. By steps 8 and 9, every constructible number lies in $T$.\n11. Let $z$ be constructible. By step 10, $z \\in F_m$ for some square-root tower, and $[F_m : \\Q] = 2^e$ for some $e \\ge 0$ by step 4.\n12. So $F_m / \\Q$ is finite, hence [[algebraic|0DFYTNZ6AM]], and $z$ is algebraic over $\\Q$.\n13. Since $\\Q \\subseteq \\Q(z) \\subseteq F_m$, the tower law gives $2^e = [F_m : \\Q(z)] \\cdot [\\Q(z) : \\Q]$. So $[\\Q(z) : \\Q]$ divides $2^e$.\n14. By unique factorisation, a positive divisor of $2^e$ is $2^r$ for some $r$ with $0 \\le r \\le e$. So $[\\Q(z) : \\Q] = 2^r$.\n\n## Idea\n\nThe numbers reachable from $\\Q$ by adjoining square roots one at a time form a field closed under square roots, so they contain all constructible numbers. Each adjunction at most doubles the degree. So a constructible number sits in a field of degree $2^e$, and its own degree divides $2^e$.\n",
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