TITLE: Inequality from Chapter 5 of the book *How to Think Like a Mathematician*
QUESTION [50 upvotes]: 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 $$\frac 17 \cdot \ln(7!) < \frac 18 \cdot \ln(8!)$$

REPLY [2 votes]: The solution occurs just by doing simple calculations,
Lets start, $\sqrt[7]{7!}<\sqrt[8]{8!}$
iff $(\sqrt[7]{7!})^{7\cdot8}<(\sqrt[8]{8!})^{7\cdot8}$
iff $(7!)^{8}<(8!)^7$
iff $(7!)^8<(7!\cdot8)^7$
iff $(7!)^8<8^7\cdot(7!)^7$
iff $(7!)<8^7$
iff $1\cdot2\cdots6\cdot7<8\cdot8\cdot8\cdot8\cdot8\cdot8\cdot8$
wich is obviously true since $1<8,2<8,\ldots,7<8$