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

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