<s>[INST] Without using a calculator, is $\sqrt[8]{8!}$ or $\sqrt[9]{9!}$ greater? Which is greater between $$\sqrt[8]{8!}$$ and  $$\sqrt[9]{9!}$$?
I want to know if my proof is correct...
\begin{align}
 \sqrt[8]{8!} &< \sqrt[9]{9!} \\
 (8!)^{(1/8)} &< (9!)^{(1/9)} \\
 (8!)^{(1/8)} - (9!)^{(1/9)} &< 0 \\
 (8!)^{(9/72)} - (9!)^{8/72} &< 0 \\
 (9!)^{8/72} \left(\left( \frac{8!}{9!} \right)^{(1/72)} - 1\right) &< 0 \\
 \left(\frac{8!}{9!}\right)^{(1/72)} - 1 &< 0 \\
 \left(\frac{8!}{9!}\right)^{(1/72)} &< 1 \\
 \left(\left(\frac{8!}{9!}\right)^{(1/72)}\right)^{72} &< 1^{72} \\
 \frac{8!}{9!} < 1 \\
 \frac{1}{9} < 1 \\
\end{align}
if it is not correct how it would be?
 [/INST] I would compare it by using logarithms:
$$\ln a = \frac{1}{8}(\ln1 + ...+\ln8)$$
$$\ln b = \frac{1}{9}(\ln1 + ...+\ln8+\ln9)=\frac{8}{9} \ln a + \frac{1}{9}\ln9$$
$$\ln b-\ln a=\frac{1}{9}(\ln9-\ln a)=\frac{1}{9} \ln\frac{9}{a}$$
$a$ is obviously less than 8 so:
$$\frac{9}{a} \gt 1$$ 
$$\ln\frac{9}{a}\gt0$$
$$\ln b - \ln a \gt 0$$
$$b \gt a$$ 
</s>