Prove $(1 +\frac{ 1}{n}) ^ {n} \ge 2$ Using induction, I proved the base case and then proceeded to prove: $$(1 + \frac{1}{n+1}) ^ {n+1} \ge 2$$ given $$(1 + \frac{1}{n}) ^ n \ge 2$$ However, I'm stuck at this point and have no clue how to go about it. Other than induction, I tried simple algebraic transformations but couldn't prove this inequality. Any pointers on how to prove this will be appreciated. 

[PS: This is my first question on stackexchange, so I'm sorry if there's anything wrong with this post and will be happy to edit if needed].
Hint Prove by induction the more general statement:

If $x &gt;0$ then 
$$(1+x)^n \geq 1+nx $$
Assuming $\left ( 1+\dfrac{1}{n}\right)^n \geq 2$, we want to prove $\left ( 1+\dfrac{1}{n+1} \right)^{n+1}\geq 2$. 

You may start by saying: $\left ( 1+\dfrac{1}{n+1}\right)^{n+1} = \left ( 1+\dfrac{1}{n+1} \right)^n\left ( 1+\dfrac{1}{n+1}\right).$

But if $\left( 1+\dfrac{1}{n}\right)^n\geq 2, $ then $\left( 1+\dfrac{1}{n+1}\right)^n\geq 2$. 

$\left( 1+\dfrac{1}{n+1}\right)^n\left ( 1+\dfrac{1}{n+1}\right)\geq 2\left ( 1+\dfrac{1}{n+1}\right)$. 

It is easy to see that $\left ( 1+\dfrac{1}{n+1}\right) \geq 1$, therefore conclusion follows.