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

$$ \begin{array}{ccl} (1+\frac{1}{n})^n &=& \sum_{k=0}^n {n \choose k}\frac{1}{n^k}\\ &=& 1 + \sum_{k=1}^n \frac{n(n-1)\cdots(n-k+1)}{k!n^k}\\ &<& \sum_{k=0}^n \frac{1}{k!}\\ &=& 2 + \sum_{k=2}^n \frac{1}{k!}\\ &<& 2 + \sum_{k=2}^n \left(\frac{1}{k-1}- \frac{1}{k}\right)\\ &=& 3 - \frac{1}{n}\\ &<& 3\\ &\le& n. \end{array} $$

At this point, the proof is completed.