TITLE: Closed form for $\prod_{n=0}^\infty (1-z^{2^n})$
QUESTION [7 upvotes]: Is there a closed form for the product $$f(z) = \prod_{n=0}^\infty 1-z^{2^n}$$
either as a formal power series or as an analytic function in the disk $|z| < 1$? It's not hard to see that Taylor series coefficients of $f$ about 0 are all $\pm 1$: 
$$f(z) = 1 - z - z^2 + z^3 - z^4 + z^5 + z^6 - z^7 + \dotsb$$
and that they form a pattern in blocks, so to speak: 
$$+- \quad\to\quad +--+ \quad\to\quad +--+-++- \quad\to\quad \dotsb$$
But I don't know much else. I came across this infinite product when I incorrectly transcribed an exercise from a book, wasting a good deal of time before returning to the book to ask it, "are you sure?". (The exercise was to prove that $\prod_{n=0}^\infty 1 + z^{2^n} = (1-z)^{-1}$, which is easy.)

REPLY [5 votes]: Converting my comment from 2016 into an answer, with elaborations:
The coefficients of $z^n$ in $f(z)$ is $-1$ to the power of the number of $1$s in the binary expansion of $n$; this is a slight variant of the Thue-Morse sequence with $1$ and $-1$ instead of $0$ and $1$ (A106400 on the OEIS). The coefficient is $1$ if $n$ is "evil" (has an even number of $1$s; A001969 on the OEIS) and $-1$ if $n$ is "odious" (has an odd number of $1$s; A000069 on the OEIS). The reciprocal
$$\frac{1}{f(z)} = \prod_{n \ge 0} \frac{1}{1 - z^{2^{n}}}$$
is the generating function for the sequence $c_n$ counting the number of partitions of $n$ into powers of $2$, which is A018819 on the OEIS. $c_{2n}$ is A000123 on the OEIS because $c_{2n+1} = c_{2n}$.
According to Kachi and Tzermias' On the $m$-ary partition numbers this problem was first studied by Euler. Mahler showed in On a special functional equation that
$$c_n = e^{O(1)} 2^{ -{k \choose 2}} \frac{n^k}{k!}$$
where $k$ is the unique positive integer satisfying $2^{k-1} k \le n < 2^k (k+1)$. This gives $k \approx \frac{W(n \log 2)}{\log 2}$ where $W(x) \sim \log x$ is the Lambert W function, which lets us write down an asymptotic for the logarithm of $c_n$ the leading term of which is
$$\log c_n \sim \frac{(\log n)^2}{2 \log 2}.$$
de Brujin's On Mahler's partition problem analyzes the asymptotics more precisely.
These asymptotics imply that $c_n$ grows faster than polynomially but slower than exponentially and so rule out $f(z)$ having a particularly simple closed form; for example it follows that $f(z)$ cannot be rational (although this has an easier proof) or more generally meromorphic or algebraic.