<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] This step doesn't look right to me:
\begin{gather}
(8!)^{(9/72)} - (9!)^{8/72} < 0 \\[6px]
(9!)^{8/72} \left(\left( \frac{8!}{9!} \right)^{(1/72)} - 1\right) < 0
\end{gather}
When you divide $(8!)^{9/72} $ by $(9!)^{8/72}$, you should get 
$$
\frac{(8!)^{9/72}}{(9!)^{8/72}} =
\frac{(8!)^{9/72}}{(9\cdot8!)^{8/72}} =
\frac{(8!)^{1/72}}{(9)^{8/72}} =
\frac{(8!)^{1/72}}{(9)^{1/9}} =
\left( \frac{(8!)^{(1/8)}}{9} \right)^{1/9}
$$
Also, note that your proof is basically of this form:
\begin{gather}
a < b \\[6px]
a - b < 0 \\[6px]
\left(\frac{a}{b} -1\right) < 0\\[6px]
\frac{a}{b} < 1
\end{gather}
You can skip several steps and just do
\begin{gather}
a < b \\[6px]
\frac{a}{b} < 1
\end{gather}
</s>