3. [4] Find the smallest positive integer $n$ such that $\underbrace{2^{2 \cdot}}_{n}>3^{3^{3^{3}}}$. (The notation $\underbrace{2^{2 \cdot \cdot}}_{n}$ is used to denote a power tower with $n 2$ 's. For example, $\underbrace{2^{2^{2}}}_{n}$ with $n=4$ would equal $2^{2^{2^{2}}}$.)
Answer: 6 Clearly, $n \geq 5$. When we take $n=5$, we have
$$
2^{2^{2^{2^{2}}}}=2^{2^{16}}4^{4^{27}}>3^{3^{27}}=3^{3^{3^{3}}} \text {. }
$$

Our answer is thus $n=6$.