Comparing $2^{F_n}$ and $2^{\varphi^n}$
Hint: We have $F_n = [\varphi^n-(-\varphi)^{-n}]/\sqrt{5} = \varphi^n/\sqrt{5} \pm o(1)$, since $|1/\varphi| &lt; 1$. Hence
$$ 2^{F_n} = 2^{\varphi^n/\sqrt{5}} 2^{\pm o(1)} = 2^{\varphi^n/\sqrt{5}} (1\pm o(1)). $$
In particular, $2^{F_n} = \Theta(2^{\varphi^n/\sqrt{5}})$. I'll let you compare $2^{\varphi^n/\sqrt{5}}$ and $2^{\varphi^n}$ yourself.