Being Inspired is Awesome!

Inspiration $\dfrac{1}{\displaystyle\prod_{n=1}^{5}\left(\sqrt[2^n]{m+2}+\sqrt[2^n]{m}\right)}$ $=\dfrac{\color{#D61F06}{(\sqrt[2^5]{m+2}-\sqrt[2^5]{m})}}{\left(\displaystyle\prod_{n=1}^{4}\left(\sqrt[2^n]{m+2}+\sqrt[2^n]{m}\right)\right)\underbrace{\left((\sqrt[2^5]{m+2}+\sqrt[2^5]{m})(\color{#D61F06}{\sqrt[2^5]{m+2}-\sqrt[2^5]{m})}\right)}_{\large(\sqrt[2^4]{m+2}-\sqrt[2^4]{m})}}$ Using $\color{teal}{(a+b)(a-b)=a^2-b^2}$ repeatedly we get: $\large\color{#0C6AC7}{\mathfrak{N}}=\displaystyle\sum_{n=0}^{2^{2^5}-1}\left(\sqrt[2^5]{m+2}-\sqrt[2^5]{m} \right)$ $\mathbf{(A~Telescopic~Series)}$ $=-0-1+\sqrt[2^5]{2^{2^5}}+\sqrt[2^5]{2^{2^5}+1}$ $=1+\sqrt[2^5]{2^{2^5}+1}$ $\large\log_2\left(\log_{2}\left((\color{#0C6AC7}{\mathfrak{N}}-1)^{2^5}-1\right)\right)+4$ $\Large =\log_2\left(\log_2\left(2^{2^5}\right)\right)+4$ $\huge =5+4=\boxed 9$