Proofs of basilar powers identities

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We all know that a simple and intuitive way to show what $2^n$ is (for $n$ an integer number) is to write it as

$$2^n = \underbrace{2\times 2\times 2\times \cdots \times 2}_{n\ \text{times}}$$

My question is: what is the most intuitive method to show the meaning of these two identities?

$$2^{-n} = \frac{1}{2^n}$$ $$2^{\frac{1}{n}} = \sqrt[n]{2}$$

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The starting point to justify (not really a proof) the two formulas in OP is to start from a good definition of the positive integer power of a number $a$. The better that I know is a

recursive definition: $$ a^0=1 \qquad a^n=a \times a^{n-1} \qquad a \in \mathbb{R} \quad n \in \mathbb{N} $$

from this definition we can easily see that this exponential function has the property: $$ (1) \qquad \qquad a^{n+m}=a^na^m $$ that implies $$ (2) \qquad \qquad (a^n)^m=a^{n\times m} $$ now we want to extend the definition to exponents in $\mathbb{Q}$ in such a way that the properties $(1)$ and $(2)$ are always true. So, we have: $$ a^0=a^{n+(-n)}=a^na^{-n}=1 $$ and this means that $a^{-n}$ have to be inverse of $a^n$.

Also, if $t=\dfrac{m}{n} \in \mathbb{Q}$, we have:

$$ a^{n \times t}=a^m \iff (a^t)^n=a^m $$ and this means, by definition of the radical, that $$ a^t=a^{\frac{m}{n}}=\sqrt[n]{a^m} $$

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By the first relation

$$2^{n}=\underbrace{2\times 2\times 2\times \cdots \times 2}_{n\ \text{times}}=\underbrace{2\times 2\times 2\times \cdots \times 2}_{n-1\ \text{times}}\times2=2^{n-1}\times2.$$

Then $$2^{n-1}=\frac{2^n}2,\\ 2^{n-2}=\frac{2^n}{2^2},\\ \cdots\\ 2^{n-m}=\frac{2^n}{2^m}.$$

By the first relation $$2^{nm} = \underbrace{2\times 2\times 2\times \cdots \times 2}_{nm\ \text{times}}=\underbrace{\underbrace{2\times 2\times 2\times \cdots \times 2}_{n\ \text{times}}\times \underbrace{2\times 2\times 2\times \cdots \times 2}_{n\ \text{times}}\cdots \underbrace{2\times 2\times 2\times \cdots \times 2}_{n\ \text{times}}}_{m\text{ times}}\\ =(2^n)^m$$ and

$$\sqrt[m]{2^{nm}}=2^n=2^{nm/m},$$ or $$\sqrt[m]{2^p}=2^{p/m}.$$