Source: Kosovo MO 2020 Grade 9, Problem 1

$2^{2^{2^{2^2}}}=2^{2^{16}}$ and $3^{3^{3^3}}=3^{3^{27}}$. Since $3^{27}>2^{16}$ and $3>2$, we must have $3^{3^{3^3}}>2^{2^{2^{2^2}}}$.

$2^{2^{2^{2^2}}}=2^{2^{16}}$ $3^{3^{3^3}}=3^{3^{27}}$ $3^{3^{27}}>2^{2^{16}} \Rightarrow 3^{3^{3^3}}>2^{2^{2^{2^2}}}$ $\blacksquare$