If $y=(1+x)\left(1+x^2\right)\left(1+x^4\right) ....\left(1+x^{2^n}\right)$ then $\frac{d y}{d x}$ at x = 0 is

1

$y=(1+x)\left(1+x^2\right)\left(1+x^4\right) .... \left(1+x^{2^n}\right)$

Multiplying numerator and denominator by (1 - x)

$\Rightarrow y=\frac{(1-x)(1+x)\left(1+x^2\right) .... \left(1+x^{2 n}\right)}{(1-x)}$

$\Rightarrow y=\left(\frac{1-x^{2^{n+1}}}{(1-x)}\right)$

∴ $\frac{d y}{d x}=\frac{(1-x) .\left\{-2^{n+1} . x^{2^{n+1-1}}\right\}-\left(1-x^{2^{n+1}}\right)(-1)}{(1-x)^2}$

So  $\left.\frac{d y}{d x}\right|_{x=0}=\frac{-2^{n+1} . 0 . 1+1-0}{1^2}=1$

Hence (1) is correct answer.