Prove increasing sequence I need to prove that $$\left(1-\dfrac{1}{2^n}\right)^n\ge \dfrac12$$  for $n\ge1$.
I thought I would prove this by showing that this is an increasing sequence with a base case at $n=1: (1-1/2^1)^1=1/2$, but I can't figure it out.
Ideas?
 A: HINT:$$1-(\dfrac12)^n=\dfrac12+\dfrac14+\dfrac18+\cdots+\dfrac{1}{2^n}$$ (easy to prove)
now put into inequality 
$$(1-\dfrac{1}{2^n})^n\ge \dfrac12\\(\dfrac12+\dfrac14+\dfrac18+\cdots+\dfrac{1}{2^n})^n \geq \dfrac12\\
\dfrac12+\dfrac14+\dfrac18+\cdots+\dfrac{1}{2^n}\geq \dfrac{1}{\sqrt[n]{2}}$$ and go for induction 
A: The tedious way when you don't have any good tricks up your sleeve:
let $f:x\mapsto \left(1-\frac{1}{2^x} \right)^x$.
Note that $f'(x)=\left(1-\frac{1}{2^x} \right)^x \left(\ln\left(1-\frac{1}{2^x} \right) + \ln 2 \dfrac{x2^{-x}}{1-2^{-x}}\right)$
Let $g:x\mapsto \ln\left(1-\frac{1}{2^x} \right) + \ln 2 \dfrac{x2^{-x}}{1-2^{-x}}$.
Note that $g'(x)=-\dfrac{\ln (2) }{\left(2^x-1\right)^2}\left(2^x x \ln (2)-2 \left(2^x-1\right)\right)$
Let $h:x\mapsto 2^x x \ln (2)-2 \left(2^x-1\right)$
Note that $h'(x)=2^x \ln (2) (x \ln (2)-1)$ which is $\geq 0$ iff $x\geq \frac{1}{\ln 2}$. So $h$ decreases over $[1,\frac{1}{\ln 2}]$ and increases over $[\frac{1}{\ln 2},\infty)$. Note that $h(1)<0$, so there is some $\alpha$ such that $h(x)\geq 0 \iff x\geq \alpha$. Therefore, $g'(x)\geq 0 \iff x\in [1,\alpha]$. So $g$ increases over $[1,\alpha]$ and decreases over $[\alpha,\infty)$. Note that $g(1)=0$ and $\lim_{x\to \infty} g(x)=0$, so $g(x)\geq 0$ whenever $x\geq 1$.
That implies $f'(x)\geq 0$ whenever $x\geq 1$, so $f$ increases, and so does the sequence $\left(1-\dfrac{1}{2^n}\right)^n$.
