Induction Proof ${n}\geq 3, (1+\frac{1}{n})^{n}\lt n$

Question

For each natural number $n$ with ${n}\geq 3, (1+\frac{1}{n})^{n}\lt n$.

So far, I've tested the claim for $n=3$, which was true. Then, I stated that I need to show that $(1+\frac{1}{n+1})^{n+1} \lt n+1$.

So first, I considered that $(1+\frac{1}{n+1})^{n+1}=(1+\frac{1}{n+1})^{n}(1+\frac{1}{n+1})$. This is also $\lt (n+1)(1+\frac{1}{n+1})$ (both sides were multipled by $(1+\frac{1}{n+1})$. I tried to combine the $(1+\frac{1}{n+1})$ with a common denominator, making it $(\frac{n+2}{n+1})$.

So basically, I have \begin{align} $$(1+\frac{1}{n+1})^{n+1}&= (1+\frac{1}{n+1})^{n}(1+\frac{1}{n+1})\lt (n+1)(1+\frac{1}{n+1})\\ &\implies (\frac{n+2}{n+1})(1+\frac{1}{n+1})^{n} \lt (n+1)(\frac{n+2}{n+1}). \end{align} At this point though, I'm lost. I know that I have to show that $(n+1)(\frac{n+2}{n+1})\lt n+1$ but from the work I've done, it isn't true because the $(n+1)$ factors cancel each other out and you are left with $(n+2)$ on the left side of the inequality.

Answer 1

For the induction step you may proceed as follows (IH stands for induction hypothesis):

$$(1+\frac{1}{n+1})^{n+1}= (1+\frac{1}{n+1})(1+\frac{1}{n+1})^n $$ $$< (1+\frac{1}{n+1})(1+\frac{1}{n})^n \stackrel{\mbox{IH}}{<}(1+\frac{1}{n+1}) \cdot n= n+\frac{n}{n+1}<n+1$$

Answer 2

$\Bigr(1+\frac{1}{n}\Bigr)^{n}\lt n \implies \Bigr(\frac{n+1}{n}\Bigr)^{n}\lt n \implies \frac{(n+1)^n}{n^n}\lt n \implies (n+1)^n \lt n^n\cdot n \implies (n+1)^n\lt n^{n+1}$

Can you continue from here?