<s>[INST] Proofs of basilar powers identities 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?
$$2^{-n} = \frac{1}{2^n}$$
$$2^{\frac{1}{n}} = \sqrt[n]{2}$$
 [/INST] 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
$$\sqrt[m]{2^p}=2^{p/m}.$$
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