Is every radical a $p^{\text{th}}$ root for some prime $p$?

Question

In case anyone wants contexts, see the bottom two lines of page 24 of http://www.math.jhu.edu/~smahanta/Teaching/Spring10/Stillwell.pdf

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The author is claiming that

"First, we can assume that each radical $a_i$ adjoined is a $p$-th root for some prime $p$.

E.g., instead of adjoining $\sqrt[6]a$ we can adjoin first $\sqrt a = b$, then $\sqrt[3]b$."

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Wouldn't the author's claim mean that each number that can be written as $x^{y}$| $x,y \in \mathbb{Q}$ is equal to $\sqrt[p]p $ for some prime $p$? I also don't understand how is it that his example is explaining this.

Could someone provide a better explanation/proof of why this is true?

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Any help/ thoughts would be really appreciated!

Answer 1

Suppose I want to solve the equation $a^r=b$ corresponding to the radical $a=\sqrt[r] b$.

Let $r=pq$ where $p$ is a prime. Then I can first solve $c^p=b$ corresponding to $c=\sqrt [p] b$ and I am left with $a^q=c$ where the exponent has fewer prime factors. I can go on adding prime roots until the exponent is reduced to $1$.

This corresponds to the fact that the original exponent $r$ can be expressed as a product of primes.