Introduction

We give a proof that

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

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

Proof

It is equivalent to show that

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

or

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

If $n\ge 3$, then

$$\begin{array}{ccl}(1+\frac{1}{n}{)}^n& =& \sum _{k=0}^n(\genfrac{}{}{0ex}{}{n}{k})\frac{1}{n^k}\\ & =& 1+\sum _{k=1}^n\frac{n(n-1)\cdots (n-k+1)}{k!n^k}\\ & <& \sum _{k=0}^n\frac{1}{k!}\\ & =& 2+\sum _{k=2}^n\frac{1}{k!}\\ & <& 2+\sum _{k=2}^n(\frac{1}{k-1}-\frac{1}{k})\\ & =& 3-\frac{1}{n}\\ & <& 3\\ & \le & n.\end{array}$$

At this point, the proof is completed.