1. $[\mathbf{3}]$ If $4^{4^{4}}=\sqrt[128]{2^{2^{2^{n}}}}$, find $n$.
Answer: $\square$ We rewrite the left hand side as
$$
\left(2^{2}\right)^{4^{4}}=2^{2 \cdot 4^{4}}=2^{2^{9}} \text {, }
$$
and the right hand side as
$$
\left(2^{2^{2^{n}}}\right)^{\frac{1}{128}}=2^{2^{\left(2^{n}-7\right)}} \text {. }
$$

Equating, we find $2^{n}-7=9$, yielding $n=4$.