Show that $\sqrt[7]{7!}<\sqrt[8]{8!}$ [duplicate]

This is from the book How to think like a Mathematician,

How can I prove the inequality $$\sqrt[\large 7]{7!} < \sqrt[\large 8]{8!}$$

without complicated calculus? I tried and finally obtained just $$\frac 17 \cdot \ln(7!) < \frac 18 \cdot \ln(8!)$$

Your inequality is equivalent to $$(7!)^8 < (8!)^7$$ divide it by $(7!)^7$, and get $$7! < 8^7$$ and this is clear, since $$1 \cdots 7 < 8 \cdots 8$$

Think of

$${\ln(7!)\over7}={\ln(1)+\cdots+\ln(7)\over7}$$

as the average of seven numbers and

$${\ln(8!)\over8}={\ln(1)+\cdots+\ln(8)\over8}$$

as the average when an eighth number is added. Since the new number is larger than the previous seven, the average must also be larger. (E.g., if you get a better score on your final than on any of your midterms, your grade should go up, not down.)