TITLE: Prove nth root of n! is less than n+1 th root of ((n+1) !): $\sqrt[n]{n!}\lt \sqrt[n+1]{(n+1)!}$?
QUESTION [6 upvotes]: Prove the inequality $$\sqrt[n]{n!}\lt \sqrt[n+1]{(n+1)!}$$

REPLY [5 votes]: Start with the pretty easy fact that the average of the logarithms of the first $n$ integers is less than the logarithm of $n+1$:
$$
\frac1n\sum_{k=1}^n\log(k)\lt\log(n+1)\tag{1}
$$
Divide by $n+1$
$$
\frac1{n(n+1)}\sum_{k=1}^n\log(k)\lt\frac1{n+1}\log(n+1)\tag{2}
$$
Add $\frac1{n+1}\sum\limits_{k=1}^n\log(k)$ to both sides
$$
\frac1n\sum_{k=1}^n\log(k)\lt\frac1{n+1}\sum_{k=1}^{n+1}\log(k)\tag{3}
$$
And $(3)$ is the logarithm of the desired inequality.