Another Proof of $(n+1)^n < n^{n+1}$

Introduction

Inspired by Sikiric’s post (URL: https://qr.ae/psOuMB), I write down another proof of

$$ (n+1)^n < n^{n+1}, $$

where $n$ is a positive integer and $n\ge 3$.

Proof

It’s equivalent to show that

$$ n\ln(n+1) < (n+1)\ln(n), $$

which is equivalent to

$$ \frac{\ln(n+1)}{n+1} < \frac{\ln(n)}{n}. $$

Let's consider function

$$ f(x)=\frac{\ln(x)}{x}. $$

Taking derivative, we have

$$ f^{\prime}(x)=\frac{1-\ln(x)}{x^2} < 0,\ \hbox{if}\ x > e. $$

If $n\ge 3$, then $n > e$. Thus,

$$ \frac{\ln(n+1)}{n+1} < \frac{\ln(n)}{n}. $$

That is,

$$ n\ln(n+1) < (n+1)\ln(n); $$

the proof is completed.