Closed form for $\prod_{n=0}^\infty (1-z^{2^n})$

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.