If $N = 2^{(2^2)}$, find real $x$ such that $N^{N^N}=2^{(2^x)}$. Understanding the solution. Harvard-MIT Math Tournament 2019 Algebra Problem:

Let $N = 2^{(2^2)}$ and $x$ be a real number such that $N^{N^N}=2^{(2^{x})}$. Find $x$.

Solution:
$$N^{(N^N)}=16^{16^{16}} = 2^{4\cdot2^{^{4\cdot2^{{4}}}}} = 2^{2^{2^{6}+2}} = 2^{2^{66}}$$
I don't understand how to get from (the previous step) $2^{4\cdot2^{^{4\cdot2^{{4}}}}}$ to (the final answer) $2^{2^{66}}$
$$(2^4)^{2^4} = 2^{4\times 2^4} = 2^{2^6}$$
$$(2^4)^{(2^4)^{2^4}}=(2^4)^{2^{2^6}}=2^{4\times 2^{2^6}}=2^{2^2 \times 2^{64}}=2^{2^{66}}$$
To my eye, the difficulty lies in the typography: It's hard to see which superscripts are at what level.* I recommend taking logarithms instead.  Using the abbreviation $\lg$ for $\log_2$, we have
$$\begin{align}
x&amp;=\lg(\lg(2^{2^x}))\\
&amp;=\lg(\lg(N^{N^N}))\\
&amp;=\lg(N^N\lg(N))\\
&amp;=\lg(N^N)+\lg(\lg(N))\\
&amp;=N\lg(N)+\lg(\lg(N))\\
&amp;=16\lg(16)+\lg(\lg(16))\\
&amp;=16\cdot4+\lg(4)\\
&amp;=64+2\\
&amp;=66
\end{align}$$
*In fact there is a typo in the OP's display: &quot;2^{4\cdot^{2^{4\cdot^{2^{4}}}}}&quot; should be &quot;2^{4\cdot{2^{4\cdot{2^{4}}}}}&quot; (without the spurious ^'s after the \cdot's), i.e
$$2^{4\cdot{2^{4\cdot{2^{4}}}}}\qquad\text{instead of}\qquad2^{4\cdot^{2^{4\cdot^{2^{4}}}}}$$
but even when TeXed correctly, I find it hard to see which superscript is at which level.