Is $\prod_{i=1}^\infty (1-\frac{1}{2^{(2^i)}})$ transcendental? Motivation. In a coin game, a player flips all their coins every turn, starting with just one coin. If the coins all land heads then the game stops; otherwise, the number of coins is doubled for the following turn. While the game may clearly terminate on any turn, there is in fact a positive probability that it will never terminate: this is the infinite product $$p = \prod_{i=1}^\infty \left(1-\frac{1}{2^{(2^i)}}\right).$$
Questions. Do we have $p \in \mathbb{R}\setminus \mathbb{Q}$? Is $p$ transcendental?
 A: Q: Is $\prod_{i=1}^\infty(1-2^{-i})$ transcendental?
The question as posed originally by the OP.A: This is the Euler function $\phi(1/2)$, see https://oeis.org/A048651 --- I don't think it is known whether it is transcendental or not [as is the case for other instances of $\phi(q)$].
A: This number is irrational. By Euler's pentagonal number theorem, we have
$$\prod_{i=1}^{\infty} (1-2^{-i}) = 1 + \sum_{k=1}^{\infty} (-1)^k \left( 2^{-k(3k+1)/2} +2^{-k(3k-1)/2} \right).$$
This shows that the binary expansion is clearly not periodic.
I have no idea whether this number is transcendental. Your number is closely related to the value of various modular forms like the Dedekind eta function at $\tau = \tfrac{\log 2}{2 \pi} i$, so $q = \tfrac{1}{2}$. Chapter 27 of Transcendental numbers, by Murty and Rath, discusses evaluating modular forms when $\tau$ is an algebraic number, but a quick search didn't show me anything about evaluating modular forms when $q$ is algebraic.
A: I can prove that the number actually described by the word problem, which is $$ \prod_{i=0}^{\infty} \left( 1- \frac{1}{2^{2^i}} \right),$$ is irrational, by a method similar to David Speyer's.
Expanding out the product, we get $$\sum_{j=0}^{\infty} \frac{(-1)^{t_j} }{2^j}= 1+\sum_{j=1}^{\infty} \frac{(-1)^{t_j} }{2^j} = \sum_{j=1}^{\infty} \frac{1}{ 2^j} + \sum_{j=1}^{\infty} \frac{(-1)^{t_j} }{2^j}  = \sum_{j=1}^{\infty} \frac{1+ (-1)^{t_j} }{2^j} = \sum_{j=1}^{\infty}  \frac{1+ (-1)^{t_j} }{2} \cdot 2^{1-j}  $$ where $t_j$ is the number of $1$s in the binary expansion of $j$.
Thus the binary digit in the $j-1$th place is $1$ if $t_j$ is even and $0$ if $t_j$ is odd.
Since this sequence is not periodic, the number is irrational.
