Comparing $e$ with $2$ using induction

Question

So, the question which came out to be challenging is to prove that $$\left(1 + {1\over n + 1}\right)^{n+1} > 2$$ for $n\in N$ by induction and not using the definition of $e$.

Answer 1

$$\left(1+\frac{1}{n+1}\right)^{n+1}\geq1+(n+1)\cdot\frac{1}{n+1}=2$$ Because by induction if $x>0$ and $(1+x)^n>1+nx$ then: $$(1+x)^{n+1}>(1+nx)(1+x)=1+(n+1)x+nx^2>1+(n+1)x$$