$\prod_{k=1}^\infty (1-1/2^k)$ converge to zero? I know this product converge
$$\prod_{k=1}^\infty (1-1/2^k),$$
but I don't know how to prove that this limite is different to zero. This is equivalent to prove that 
$$\sum^{\infty}_{k=1} \log(1-1/2^k)$$ converge. I can't get it. Could you give me any hint? Thank!
I'm trying to prove that 
$$\frac{|GL_n(\mathbb{F}_2)|}{2^{n^2}} \to \alpha,$$
where $\alpha > 0$ 

$\log(1-x)$ is a concave function on $[0,1)$, so for any $x\in[0,1/2]$ we have
$$ \log(1-x)\geq -2\log(2) x $$
immediately implying
$$ \sum_{k\geq 1}\log\left(1-\frac{1}{2^k}\right) \geq -2\log(2)\sum_{k\geq 1}\frac{1}{2^k} = -2\log(2) $$
and
$$ \prod_{k\geq 1}\left(1-\frac{1}{2^k}\right) \geq \frac{1}{4}.$$
Much better approximations can be derived through the Mellin transform, as done by Marko Riedel here.
Creative telescoping is also a chance. Numerically the LHS is $\approx 0.288788095$.
