I'm not sure how to simplify this type of exponential expression. I would like to know $k$ many $log_{2}^{k}(n)$ such that $n \leq 1$ $$n = (2^{2^{2^{15}}})^2$$ So attempt I to simplify $$n = 2^{(2^{2^{15}} ) + (2^{2^{15}})}$$ From here I'm not exactly sure how to simplify further $$n = 2^{2^{4^{30}}}$$ So I figured I could bring the exponent of exponents down $$n = 2^{16^{30}}$$ Turns out to be a very big number $$n = 2^{1.329\times 10^{36}}$$
So $k = 2^{1.329\times 10^{36}}$?
We have $$n = (2^{2^{2^{15}}})^2=2^{(2^{2^{15}}+1)}$$ The $+1$ is negligible. If you take one log, you have $$\log_2(n)=2^{2^{15}}\\\log_2(\log_2(n))=2^{15}\\$$ A third one gets you to $15$, then a little less than $4$, then a little less than $2$, then less than $1$. So it takes six applications of the log function.