I have to show that $$\sqrt[7]{7!}<\sqrt[8]{8!}$$ I've already tried raising both sides to $7\cdot8$ and got $$(7!)^8<(8!)^7$$ and, after a few computations, I obtained $$7!<8^7$$ What should I do to prove the above result? Should I use another approach from the beginning?
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2Does this answer your question? Inequality from Chapter 5 of the book *How to Think Like a Mathematician* – Alexander51413 Dec 08 '21 at 15:57