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?
Best Answer
$$(\sqrt[8]{8!})^ {72}= (8!)^9 = (8!) (8!)^8 $$
$$(\sqrt[9]{9!})^ {72} = (9!)^8 = (9\times 8!)^8 = 9^8 (8!)^8$$
The second one, wins hands down.