I know that
$$\prod_1^\infty \left(1-\frac{1}{2^n}\right)$$
converges to a positive number because the series $\sum 2^{-n}$ is convergent. Do we know the limit? If so, how?
Aside: I am interested in this product because it describes asymptotically the fraction of $n\times n$ matrices with entries in $\mathbb{F}_2$ that are nonsingular.
Let $$P = \prod_{n\ge 1} \left(1 - \frac{1}{2^n} \right).$$ Introduce $$S = \log P = \sum_{n\ge 1} \log\left(1 - \frac{1}{2^n} \right).$$
Now recall the harmonic sum identity for Mellin transforms: $$\mathfrak{M}\left(\sum_{k\ge 1} \lambda_k g(\mu_k x);s\right) = \left(\sum_{k\ge 1} \frac{\lambda_k}{\mu_k^s} \right) g^*(s)$$ where $g^*(s)$ is the Mellin transform of $g(x).$
Introducing $$S(x)= \sum_{n\ge 1} \log\left(1 - \frac{1}{2^{nx}} \right)$$ so that $S=S(1)$ we see that $S(x)$ is harmonic with parameters $$\lambda_k = 1, \quad \mu_k = k \quad\text{and}\quad g(x) = \log\left(1 - \frac{1}{2^x} \right)$$ and may be evaluated (approximated) by inverting its Mellin transform.
Now $$\mathfrak{M}\left(\log\left(1 - \frac{1}{2^x} \right);s\right) = \int_0^\infty \log\left(1 - \frac{1}{2^x} \right) x^{s-1} dx = - \int_0^\infty \sum_{q\ge 1} \frac{2^{-qx}}{q} x^{s-1} dx.$$ This is $$-\frac{\Gamma(s)}{(\log 2)^s} \sum_{q\ge 1} \frac{1}{q} \frac{1}{q^s} = -\frac{\Gamma(s)}{(\log 2)^s} \zeta(s+1).$$ It follows that the Mellin transform of $S(x)$ is $$Q(s) = \mathfrak{M}(S(x); s) = -\frac{1}{(\log 2)^s}\Gamma(s)\zeta(s)\zeta(s+1).$$ Now perform Mellin inversion with the inversion integral being $$\frac{1}{2\pi i} \int_{3/2-i\infty}^{3/2+i\infty} Q(s)/x^s ds$$ and shifting the integral to the left for an expansion about zero. There are only three singularities to consider because the trivial zeros of the two zeta function terms cancel the poles of the gamma function.
We have $$\operatorname{Res}(Q(s)/x^s; s=1) = -1/6\,{\frac {{\pi }^{2}}{\log 2\times x}},$$ $$\operatorname{Res}(Q(s)/x^s; s=0) = 1/2\,\log \left( 2\,\pi \right) -1/2\,\log \log 2 -1/2\,\log x$$ and $$\operatorname{Res}(Q(s)/x^s; s=-1) = 1/24\,\log 2 \times x.$$ Putting $x=1$ we obtain that $$S(1) = S \approx -1/6\,{\frac {{\pi }^{2}}{\log 2}} + 1/2\,\log \left( 2\,\pi \right) -1/2\,\log \log 2 + 1/24\,\log 2.$$ This approximation is excellent (good to $24$ digits) but not quite exact. E.g. setting $x=1/2$ which is closer to zero we get $50$ good digits, setting $x=1/5$ we get $123$ good digits and so on.
The conclusion is that $$ P \approx e^{-\pi^2/6/\log 2} \times \sqrt{2\pi} \times \frac{2^{1/24}}{\sqrt{\log 2}} \approx 0.2887880950866024212788997.$$
Adddendum, July 2022. We can actually do a somewhat better and compute a functional equation for $S(x).$ This requires an evaluation of the remainder, which is
$$\frac{1}{2\pi i} \int_{-3/2-i\infty}^{-3/2+i\infty} Q(s)/x^s \; ds.$$
Substitute $s = - t$ in this integral to get (we get one minus from $Q(s)$, another from the differential, and a third reversing the direction of the line)
$$-\frac{1}{2\pi i} \int_{3/2-i\infty}^{3/2+i\infty} (\log 2)^t \Gamma(-t) \zeta(-t) \zeta(1-t) x^{t} dt$$
In view of the desired functional equation we now use the functional equation of the Riemann zeta function on $Q(s)$ to prove that the integrand of the last integral is in fact a scaled version of $Q(t).$
Start with the functional equation $$\zeta(1-s) = \frac{2}{2^s\pi^s} \cos\left(\frac{\pi s}{2}\right) \Gamma(s) \zeta(s)$$ and substitute this into $Q(s)$ to obtain
$$Q(s) = - \frac{1}{(\log 2)^s} \frac{\zeta(1-s) 2^s \pi^s}{2\cos\left(\frac{\pi s}{2}\right)} \zeta(s+1).$$
Apply the functional equation again (this time to $\zeta(s+1)$) to get
$$- \frac{1}{(\log 2)^s} \frac{\zeta(1-s) 2^s \pi^s}{2\cos\left(\frac{\pi s}{2}\right)} 2^{s+1} \pi^s \cos\left(-\frac{\pi s}{2}\right) \Gamma(-s) \zeta(-s).$$
This is
$$Q(s) = - 2^{2s} \pi^{2s} \frac{1}{(\log 2)^s} \Gamma(-s) \zeta(-s) \zeta(1-s).$$
We thus have for the remainder integral (there was a minus in front and the multiple of $Q(t)$ brings another one)
$$\frac{1}{2\pi i} \int_{3/2-i\infty}^{3/2+i\infty} (\log 2)^t \frac{Q(t)}{2^{2t} \pi^{2t}} (\log 2)^t x^{t} dt \\ = \frac{1}{2\pi i} \int_{3/2-i\infty}^{3/2+i\infty} Q(t) \left(\frac{4\pi^2}{x(\log 2)^2}\right)^{-t} \; dt.$$
Together with the residues we have established the following functional equation
$$\bbox[5px,border:2px solid #00A000]{ \begin{gather} S(x) = - \frac{1}{6} \frac{\pi^2}{\log 2 \times x} + \frac{1}{2} \log(2\pi/\log 2/ x) + \frac{1}{24} \log 2 \times x \\ + S \left(\frac{4\pi^2}{x(\log 2)^2}\right) \end{gather}.}$$
The fixed point at $x=2\pi/\log 2$ of the recursion does not produce a value here, but we get by way of confirming the computation that
$$0 = -\frac{1}{6} \frac{\pi^2}{2\pi} + \frac{1}{24} 2\pi.$$
This is a tautology but, your product is a $q$-Pochhammer symbol $(\frac{1}{2}; \frac{1}{2})_\infty$. One connection to Jacobi-Theta functions is:
$$\left(\frac{1}{2}; \frac{1}{2}\right)_\infty=\frac{2^{1/24}}{\sqrt{3}}\vartheta_2\left(\frac{1}{6}\pi,\frac{1}{2^{1/6}}\right)$$
where $\vartheta_n$ is the Jacobi theta function.
