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?
 A: $$(\sqrt[8]{8!})^ {72}= (8!)^9 = (8!)  (8!)^8  $$
$$(\sqrt[9]{9!})^ {72} = (9!)^8 = (9\times 8!)^8 = 9^8 (8!)^8$$
The second one, wins hands down.
A: 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:
$$\frac{9}{a} \gt 1$$ 
$$\ln\frac{9}{a}\gt0$$
$$\ln b - \ln a \gt 0$$
$$b \gt a$$ 
A: Instead of comparing the diference, comparing the ratio should be easy in such cases:
$$\left(\frac{8!^{1/8}}{9!^{1/9}}\right)^{8}=\frac{8!}{9!^{8/9}}=\frac{8!}{9!}(9!)^{1/9}=\left(\frac{9!}{9^9}\right)^{1/9}<1$$
this quantity is inferior than $1$ because $9!<9^9$
if $a$ is strictly positive real and $n$ is positive rational and also $a^n<1$ then $a<1$
which says:
$$\frac{8!^{1/8}}{9!^{1/9}}<1$$ QED. 
A: For any growing sequence, the arithmetic mean of the first terms is a growing function.$^*$
By taking the logarithm, this extends to the geometric mean.

$$^*t_1,t_2,\cdots t_n<t_{n+1}\implies s_n<nt_{n+1}\implies\frac{s_n}n<\frac{s_n+t_{n+1}}{n+1}=\frac{s_{n+1}}{n+1}.$$
A: You are implicitly writing that $$\frac{(8!)^{(9/72)}}{(9!)^{(8/72)}} = \left( \frac{8!}{9!} \right)^{(1/72)}$$
which is wrong.
A: Another way to see this is to convert both sides into two "averages".


*

*"Uniform distribution on $\{\log1,\dots,\log8\}$": $$\frac{\log1 + \dots + \log8}{8}$$

*"Uniform distribution on $\{\log1,\dots,\log9\}$": $$\frac{\log1 + \dots + \log9}{9}$$


Intuitively, the later has a greater "expectation", so the result follows.  If you're not satisfied with this probabilistic interpretation, break the later into a sum of two terms and regroup them as
\begin{align}
& \frac{\log1 + \dots + \log8}{8} < \frac{\log1 + \dots + \log8}{9} + \frac{\log9}{9} \\
&\iff (\log1 + \dots + \log8) \left(\frac18-\frac19\right) < \frac{\log9}{9} \\
&\iff \frac{\log1 + \dots + \log8}{72} < \frac{\log9}{9} \\
&\iff \log1 + \dots + \log8 < 8 \log9
\end{align}
The last inequality is true since $\log$ is strictly increasing.
A: We have obviously $8!<9^8$. Hence, it follows that $(9!)^8=(8!\cdot9)^8=(8!)^8\cdot 9^8>(8!)^8\cdot(8!)=(8!)^9$. This implies that $(9!)^{1/9}>(8!)^{1/8}$.
A: 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}
A: $$(8!)^9=(8!)^8\cdot 8! < (8!)^8\cdot 9^8= (9!)^8 $$
A: You need only note that $9^8>8!,$ then we'd be done. And  $9^8>8!$ since all the eight factors on the left exceed all the eight factors on the right.
Then multiplying both sides by $(8!)^8$ gives $(8!)^89^8>(8!)^88!,$ or $$(9!)^8>(8!)^8,$$ and by taking $1/72$nd powers, we obtain the result $$(9!)^{1/9}>(8!)^{1/8}.$$
