Proof of $\sqrt{2^{2^k}} = 2^{2^{k-1}}$?

Question

It's quite easy to observe that for $k \ge 0$:

$$ \begin{align} 2^{2^k} &= 4, 16, 256, 65536, \dots\\ \sqrt{2^{2^k}} &= 2, 4, 16, 256,\dots \end{align} $$

More in general:

$$ \sqrt{2^{2^k}} = 2^{2^{k-1}} $$

How can I prove this identity?

Answer 1

$$\begin{align} \left(2^{2^{k-1}}\right)^2 & = \left(2^{2^{k-1}}\right)\left(2^{2^{k-1}}\right)\tag{1}\\ & = \left(2^{2^{k-1}+2^{k-1}}\right)\tag{2}\\ & = \left(2^{2(2^{k-1})}\right)\tag{3}\\ & = 2^{2^{k}}\tag{4}\\ \end{align}$$

To go from $(1)$ to $(2)$, we apply the rule that $a^ba^c=a^{b+c}$.

Going from $(2)$ to $(3)$, we note that $a+a = 2a$.

Going from $(3)$ to $(4)$, we again apply the rule that $a^ba^c=a^{b+c}$.

Answer 2

$$\sqrt{2^{2^k}} = \left(2^{2^k}\right)^{1/2} = 2^{2^k\cdot \frac{1}{2}} = 2^{2^{k-1}} $$

Answer 3

square the item on the right hand of '=' sign ,you will see that.

Answer 4

$\sqrt{2^{2^k}}=\sqrt{2^{2^{k+1-1}}}=\sqrt{2^{2^{(k-1)+1}}}=\sqrt{2^{2\cdot 2^{k-1}}}=\sqrt{(2^{2^{k-1}})^2}=2^{2^{k-1}}$