Prove nth root of n! is less than n+1 th root of ((n+1) !): $\sqrt[n]{n!}\lt \sqrt[n+1]{(n+1)!}$? 
Prove the inequality $$\sqrt[n]{n!}\lt \sqrt[n+1]{(n+1)!}$$ 


This is only an explanation rather than a rigorous proof, but it might be more convincing than a rigorous proof (and can be converted to one, if desired).
Note that
$$(n!)^{1/n}\lt((n+1)!)^{1/(n+1)}\
\iff{\ln1+\ln2+\cdots+\ln n\over n}\lt{\ln1+\ln2+\cdots+\ln n+\ln(n+1)\over n+1}$$
Each expression in the equivalent inequality can be interpreted as an average.  The second can be seen as an updated average, with one additional value, $\ln(n+1)$, that is larger than all the other values.  It's thus clear that this average exceeds the first.  (E.g., if your grades go up from year to year, your grade point average should not go down.)
