There is a known identity:
$$\prod_{k=0}^{\infty} \left( 1+x^{2^k} \right)=\frac{1}{1-x}, ~~~~~|x|<1$$
It's easy to derive it by converting it to a telescoping product as shown in this answer.
However, we can't use the same method here.
$$\left( 1-x^{2^k} \right) \left( 1+x^{2^k} \right)=\left( 1-x^{2^{k+1}} \right)$$
$$\left( 1-x^{2^k} \right) =\frac{\left( 1-x^{2^{k+1}} \right)}{ \left( 1+x^{2^k} \right)}$$
This product will not telescope. We can't even use this to find something new about it:
$$p(x)=\prod_{k=0}^{\infty} \left( 1-x^{2^k} \right)=\frac{\prod_{k=0}^{\infty} \left( 1-x^{2^{k+1}} \right)}{\prod_{k=0}^{\infty} \left( 1+x^{2^k} \right)}$$
We only get the obvious recurrence relation:
$$p(x)=(1-x)~p(x^2)$$
Mathematica gives this plot (for 25 terms).
Does this product have a closed form?