Theorem 1.21 in Baby Rudin: How do we obtain $\sqrt[m]{\sqrt[n]{a}}=\sqrt[mn]{a}$ for any $a>0$ and $m, n \in \mathbb{N}$?

Question

Here is Theorem 1.21 in Principles of Mathematical Analysis by Walter Rudin, 3rd edition:

For every real $x>0$ and every integer $n > 0$ there is one and only one positive real $y$ such that $y^n=x$.

This number $y$ is written as $\sqrt[n]{x}$ or $x^{1/n}$.

Immediately following the proof of this theorem is the following corollary:

If $a$ and $b$ are positive real numbers and $n$ is a positive integer, then $$(ab)^{1/n} = a^{1/n} b^{1/n}.$$

Using induction, from the above corollary, we can infer the following:

If $a$ is a positive real number and $n$ and $k$ are positive integers, then $$\sqrt[n]{a^k} = \left(\sqrt[n]{a}\right)^k.$$

And, then we can even extend the scope of the last result to include all integers $k$.

Now my question is this:

Using the machinery developed by Rudin, how do we arrive at the following result?

If $a$ is a positive real number and $m$ and $n$ are positive integers, then $$\sqrt[m]{\sqrt[n]{a}} = \sqrt[mn]{a}.$$

Answer 1

We want to show that $\sqrt[m]{\sqrt[n]{a}} = \sqrt[mn]{a} $

Let $b$ be the unique value such that $b = \sqrt[mn]{a} $. Then $b^{mn} = a $.

There is a unique value $c$ such that $c^m = a$.

There is a unique value $d$ such that $d^n = c$.

Then $d =c^{1/n} =(a^{1/m})^{1/n} $.

But $d^{mn} =(d^n)^m =c^m =a $.

Since these values are unique, $d = b$.

Answer 2

In the very first box, you have that $x^{1/m} =\sqrt[m] {x}$. Use this property twice, simplify, and use it again to put it back in radical form.

$$\sqrt [m] {\sqrt[n] a} = \sqrt[m] {a^{1/n}} = (a^{1/n})^{1/m} = a^{1/nm} =\ \sqrt[nm]{a}$$

Answer 3

For any $p,q\in\Bbb N$ we have $$(\sqrt[pq]{a^p})^q=\sqrt[pq]{a^{pq}}=a$$ so $$\sqrt[q]a=\sqrt[pq]{a^p}$$

Now, $$(\sqrt[mn]a)^m=\sqrt[mn]{a^m}=\sqrt[n]a$$ Thus, $$\sqrt[m]{\sqrt[n]a}=\sqrt[mn]a$$