TITLE: Is my proof correct for: $\sqrt[7]{7!} < \sqrt[8]{8!}$
QUESTION [11 upvotes]: I have to show that 
$$\sqrt[7]{7!} < \sqrt[8]{8!}$$
and I did the following steps
\begin{align}
 \sqrt[7]{7!} &< \sqrt[8]{8!} \\
 (7!)^{(1/7)} &< (8!)^{(1/8)} \\
 (7!)^{(1/7)} - (8!)^{(1/8)} &< 0 \\
 (7!)^{(8/56)} - (8!)^{7/56} &< 0 \\
 (8!)^{7/56} \left(\left( \frac{7!}{8!} \right)^{(1/56)} - 1\right) &< 0 \\
 \left(\frac{7!}{8!}\right)^{(1/56)} - 1 &< 0 \\
 \left(\frac{7!}{8!}\right)^{(1/56)} &< 1 \\
 \left(\left(\frac{7!}{8!}\right)^{(1/56)}\right)^{56} &< 1^{56} \\
 \frac{7!}{8!} < 1 \\
 \frac{1}{8} < 1 \\
\end{align}
Did I do this properly? Is this way the best way or is there another much easier way?
Thanks a bunch!

REPLY [16 votes]: Your proof is correct, but perhaps a bit elaborate.
How about:
\begin{align}
&&\sqrt[7]{7!} &< \sqrt[8]{8!}\\
\iff&&(7!)^8 &< (8!)^7\\
\iff&&(7!)^7 7! &< (8\cdot 7!)^7\\
\iff&&7! &< 8^7
\end{align}