Can we find a subfield $F$ of $\mathbb C$, such that ${ F}^{*p}={ F}^{*}$ for any prime $p\neq 5$ and ${F}^{*5}\neq F^{*}$? We know for field $\mathbb R$, ${\mathbb R}^{*p}={\mathbb R}^{*}$, for any prime $p\neq 2$ and ${\mathbb R}^{*2}\neq {\mathbb R}^{*}$. (${\mathbb R}^{*}$  is the multiplicative group of $\mathbb R$.) Can we find a subfield $F$ of $\mathbb C$ such that ${ F}^{*p}={ F}^{*}$ for any prime $p\neq 5$ and ${F}^{*5}\neq F^{*}$?

Here's a possible construction.
If $F$ is a countable subfield of $\mathbb{C}$, we can form the field $F'$ by splitting all polynomials of the form $X^p-a$ for $p\neq 5$ a prime and $a\in F$. There are countably many such polynomials, so order them in some way and split them successively. 
Now, iterate this procedure. Starting with $F_0=\mathbb{Q}$, define $F_{n+1}=F_n'$ and $F_\infty = \cup F_n$. Then $F_\infty$ is a subfield of $\mathbb{C}$, and any of its elements contains all of its $p$th roots for $p\neq 5$ a prime. (This is because any such element belongs to some $F_n$, so all its roots are already in $F_{n+1}$.)
I claim that no fifth root of $2$ is in $F_\infty$. (Thanks to Yuval Dor for this argument). In fact, let's even extend $F_\infty$ a little bit: in the first step we constructed $F_1$ by adding, among other things, all roots of unity except those of order 5. Let's throw in the roots of $X^5-1$ as well, so now $F_1$ actually contains all roots of unity of prime order. 
No fifth root of $2$ can belong in this extended $F_1$. If any root did, then all of them would, since $F_1$ contains the 5th roots of unity. But the splitting field of $X^5-2$ is a non-abelian extension of $\mathbb{Q}$, so it cannot be contained in a cyclotomic field. 
Finally, by Kummer theory, splitting polynomials of the form $X^p-a$ over a field that has the necessary roots of unity can only result in a trivial extension of an extension of order $p$. So the order $5$ can never show up, meaning $\sqrt[5]{2}$ can't appear. 
