The limit of infinite product $\prod_{k=1}^{\infty}(1-\frac{1}{2^k})$ is not 0? Is there any elementary proof that the limit of infinite product $\prod_{k=1}^{\infty}(1-\frac{1}{2^k})$ is not 0?

The following requires no tools other than induction. In particular, machinery from the calculus is not used.
Let $f(n)$ be the product up to the term $1-\dfrac{1}{2^n}$.
 We show by induction that 
$$f(n)\ge\frac{1}{4} +\frac{1}{2^{n+1}}.$$
 The result is true at $n=1$.  For the induction step, note that
 $$f(m+1)=f(m)\left(1-\frac{1}{2^{m+1}}\right)\ge\left(\frac{1}{4}+\frac{1}{2^{m+1}}\right)\left(1-\frac{1}{2^{m+1}}\right).$$
Expand the product on the right. We get
$$\frac{1}{4}+\frac{1}{2^{m+1}}-\frac{1}{4}\cdot\frac{1}{2^{m+1}}-\frac{1}{2^{2m+2}}.\qquad(\ast)$$
Rewrite $\dfrac{1}{4}\cdot\dfrac{1}{2^{m+1}}$ as $\dfrac{1}{2^{m+2}}-\dfrac{1}{2^{m+3}}$.
Then $(\ast)$ becomes
$$\frac{1}{4}+\frac{1}{2^{m+2}}+\frac{1}{2^{m+3}}-\frac{1}{2^{2m+2}}.$$
The term $\dfrac{1}{2^{m+3}}-\dfrac{1}{2^{2m+2}}$ is non-negative. Thus $f(m+1)\ge\dfrac{1}{4}+\dfrac{1}{2^{m+2}}$, which completes the induction step.
