Proof by induction $(1+\frac{1}{n})^n \lt n$ for all $n \ge 3$

Question

Prove by induction that $$(1+\frac{1}{n})^n \lt n$$ for all $n \ge 3$. So I have the smallest case done where $n=3$: $$(1+\frac{1}{3})^3 < 3$$ $$2.37 <3 $$ Now I have the $n+1$ case set up like this: $$(1+\frac{1}{m+1})^{m+1}<m+1$$ but I'm unsure where to go from here. Should I split up the left hand side into $$(1+\frac{1}{m+1})^m(1+\frac{1}{m+1})$$

Answer 1

Try this: Show by induction that $$\left(1 + {1\over n}\right)^n < 3 - {1\over n}$$ for $n \ge 1.$

Answer 2

$$(1+\frac{1}{m+1})^m(1+\frac{1}{m+1}) \le \color{blue}{(1+\frac{1}{m})^m}(1+\frac{1}{m+1}) \le m(1+\frac{1}{m+1})$$

Try to expand and conclude.