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?
$$2^{-n} = \frac{1}{2^n}$$
$$2^{\frac{1}{n}} = \sqrt[n]{2}$$
 A: 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
$$
a^t=a^{\frac{m}{n}}=\sqrt[n]{a^m}
$$
A: 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
$$\sqrt[m]{2^p}=2^{p/m}.$$
