<s>[INST] 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?
 [/INST] This is similar to the beginning of Alex's answer, however it proceeds differently.
We express the log of the product as a sum of logarithms
$$\small L=-\log(\prod_{k=1}^\infty (1-{1 \over 2^k})) = -\sum_{k=1}^\infty \log(1-{1 \over 2^k})$$ 
Next we expand each of the logs according to its power series and write this as a double sum; then we change order of summation and sum up columnwise; this is allowed because both directions provide convergent sums. So
$$\small \begin{array} {lllllll}
-\log(1-1/2)=&+1/2 & +1/2 \cdot  1/2^2 & +1/3  \cdot 1/2^3 &  +1/4  \cdot 1/2^4 & \ldots \\
-\log(1-1/4)=&+1/4 & +1/2 \cdot  1/4^2 & +1/3  \cdot 1/4^3 &  +1/4  \cdot 1/4^4 & \ldots \\
-\log(1-1/8)=&+1/8 & +1/2 \cdot  1/8^2 & +1/3  \cdot 1/8^3 &  +1/4  \cdot 1/8^4 & \ldots \\
-\log(1-1/16)=&+1/16 & +1/2 \cdot  1/16^2 & +1/3  \cdot 1/16^3 &  +1/4  \cdot 1/16^4 & \ldots \\
\cdots & & \cdots &\cdots &\cdots & \\
\hline \\
L  = & 1 & + 1/2 \cdot 1/3 & + 1/3 \cdot 1/7 & + 1/4 \cdot 1/15 & \ldots \\
\hline \\
\zeta(2)=&1&+ 1/2\cdot 1/2& + 1/3 \cdot 1/3 & + 1/4 \cdot 1/4 & \dots \\
\end{array}
$$
We compare L with $\small \zeta(2) $ and observe, that the absolute value each term in L is smaller than the corresponding term in $\small \zeta(2)$ thus L is a finite/converging sum $\small 0 \lt L \lt \zeta(2) \lt \infty $ and thus  $\small \exp(-L) \ne 0 $ . QED .
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