For all positive real $ x,y$, we have $ \log_{4}(x) > \log_{4}(y)$ iff $ x > y$. Take the logarithm base-4 of both sides: $ 4^{4^{4^{4^{4}}}} \ ? \ 5^{5^{5^{5}}}$ $ \log_{4}\left(4^{4^{4^{4^{4}}}}\right) \ ? \ \log_{4}\left(5^{5^{5^{5}}}\right)$ $ 4^{4^{4^{4}}} \ ? \ 5^{5^{5}} \log_{4}(5)$ Repeat: $ \log_{4}\left(4^{4^{4^{4}}}\right) \ ? \log_{4}\left(\ 5^{5^{5}} \log_{4}(5)\right)$ $ 4^{4^{4}} \ ? \ 5^{5}\log_{4}(5) + \log_{4}(\log_{4}(5))$ Since $ 5 < 4^2$, $ \log_{4}(5) < 2$. Also, $ \log_{4}(\log_{4}(5)) < \log_{4}(2) < 1$. Thus, $ 5^{5}\log_{4}(5) + \log_{4}(\log_{4}(5)) < 3125(2)+1 < 6251$. However, $ 4^{4^4} = 4^{256} = 4^7 \cdot 4^{249} = 16384 \cdot 4^{249} > 16384$. Therefore: $ 4^{4^{4}} \ > \ 5^{5}\log_{4}(5) + \log_{4}(\log_{4}(5))$. And thus: $ 4^{4^{4^{4^{4}}}} \ > \ 5^{5^{5^{5}}}$.

$[x]_n:=x^{x^{...^x}}$, $n$-times, where $n\geq1$ From the definition $[x]_{n+1}=x^{[x]_n}$ Let's use induction on $n$ to prove the statement: $\frac{[4]_{n+1}}{[5]_n}>2$ Basis: $n=1$, $\frac{[4]_2}{[5]_1}=\frac{4^4}{5}=\frac{256}{5}>2$ Inductive step: $\frac{[4]_{n+1}}{[5]_n}=\frac{4^{[4]_n}}{5^{[5]_{n-1}}}=\frac{(4^{\frac{[4]_n}{[5]_{n-1}}})^{[5]_{n-1}}}{5^{[5]_{n-1}}}>(\frac{16}{5})^{[5]_{n-1}}>3^{[5]_{n-1}}>2$ Therefore, $\frac{[4]_{5}}{[5]_4}>2\iff [4]_5>2\cdot[5]_4>[5]_4\iff4^{4^{4^{4^4}}}>5^{5^{5^5}}$