Proof of $(\sqrt[n]{a})^n=(\sqrt[n]{a^n})$

Question

Definition: Given a natural number greater than or equal to $1$ and a real number $a\geq 0$, we call the n-th root of $a$ and write $\sqrt[n]{a}$ that nonnegative real number $b$ such that $b^n=a$, i.e. $b=\sqrt[n]{a} \iff b^n=a$.

Now from the definition, is easy to see that $$(\sqrt[n]{a})^n=a $$

My question is how do I prove the equality (in red color)?

$$\large(\sqrt[n]{a})^n=\color{red}{(\sqrt[n]{a^n})}$$

Answer 1

Clearly, $\sqrt[n]{a^n} = a$ suffices. Let $b = \sqrt[n]{a^n}$. Then we have $b^n = a^n$ by the stated definition. To reach the goal $a = b$, we need injectivity (for positive arguments) of the power function $a \mapsto a^n$.

But this follows from the fact that this function is monotonically increasing, which holds since multiplication with a positive number is monotonically increasing as given by the ordered field axioms.

Answer 2

We fix integer $n \geqslant 1$ and $a \geqslant 0,$ real. Your definition implicitly assumes the existence and uniqueness of $\sqrt[n]{a}\;(\geqslant0)$, which needs proof. Despite this, I think we can safely skip such a proof for your audience.

Why not proceed as follows? (The uniqueness is essential and should be pointed out.)

1. Prove that $\sqrt[n]{xy} = \sqrt[n]{x} \sqrt[n]{y},\;$ for $x \geqslant 0$ and $y \geqslant 0$.

2. By induction (or skipping induction and appealing to plausibility), prove that for any fixed $k$, $\sqrt[n]{x_1 x_2 \cdots x_k} = \sqrt[n]{x_1} \sqrt[n]{x_2} \cdots \sqrt[n]{x_k}.\;$ (Note you will need to assume that a product of non-negatives is again non-negative, for the left-hand side to exist.)

3. Now with $k=n,$ we write $\sqrt[n]{a^n}= \sqrt[n]{a} \sqrt[n]{a} \cdots \sqrt[n]{a}.$

See that $\sqrt[n]{a} \sqrt[n]{a} \cdots \sqrt[n]{a}$ is the definition of $(\sqrt[n]{a})^n$.