Is a field $K$ real closed if it is closed under $n$-th roots?

Question

A field $K$ is real closed if the following two conditions hold:

1. for every $a \in K$ there is $b \in K$ such that $a = b^2$ or $a = -b^2$, and
2. every polynomial of odd degree has a root in $K$.

Now let $K$ be an ordered field such that $a^{\frac{1}{n}} \in K$ for every $a \in K$ with $a > 0$ and natural number $n \in \mathbb{N}$. Does it follow that $K$ is real closed?

Answer 1

Asking that every polynomial of odd degree has a root in $K$ is much stronger than asking that $K$ is closed under all $n$-th roots of positive elements. Recall the famous theorem of Galois that there are polynomials of degree $5$ and higher over $\mathbb{Q}$ whose roots can't be expressed using the field operations and $n$th roots...

Explicitly, let $K$ be the ordered field obtained by starting with $\mathbb{Q}$ and iteratively adjoining all real $n$-th roots of positive elements. $K$ is closed under $n$-th roots of positive elements, but it is not real closed because, for example, the polynomial $x^5-x-1$ does not have a root in $K$.