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?

Solution 1:

$$(\sqrt[8]{8!})^ {72}= (8!)^9 = (8!) (8!)^8 $$

$$(\sqrt[9]{9!})^ {72} = (9!)^8 = (9\times 8!)^8 = 9^8 (8!)^8$$

The second one, wins hands down.

Solution 2:

You are implicitly writing that $$\frac{(8!)^{(9/72)}}{(9!)^{(8/72)}} = \left( \frac{8!}{9!} \right)^{(1/72)}$$ which is wrong.