Lower bounding $(1-2^{-n})^n$

Question

I'm trying to prove the following statement:

For each $n \in \mathbb{N}$, $(1-2^{-n})^n \geq \frac{1}{2}$

My attempts:

I've first written $(1-2^{-n})^n=(\frac{2^n - 1}{2^n})^n=\frac{(2^n - 1)^n}{2^{n^2}}$. I've tried using the binomial theorem and lower bounding the numerator, or writing the numerator as $(1+2+...+2^{n-1})$, but it hasn't worked. I've also tried analyzing the real-valued function $(1-2^{-x})^x$, but its derivatives are too complicated to analyze.

What else can I do in order to prove the statement?

Thanks

Answer 1

By Bernoulli's inequality, $$(1-2^{-n})^n\ge 1-n2^{-n},$$ so all one needs is to prove that $$n2^{-n}\le\frac12,$$ equivalently $$n\le 2^{n-1}$$ etc.