Is my proof correct for: $\sqrt[7]{7!} < \sqrt[8]{8!}$
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}