Let  $y\left(x\right)=(1+x)(1+x^2)(1+x^4)(1+x^8)(1+x^{16})$  then  $y^1-y^{11}$  at   $x=-1$ is equal to

$$\begin{gathered} y=\frac{x^{32}-1}{x-1} y(x-1)=x^{32}-1 y(-1)=0,and{y}^1(x-1)+y=32x^{31} \\ \Rightarrow y^1(-1)=16 y^{11}=(x-1)+y^1+y^1=32\times 31x^{30} \\ at x=-1 \\ \Rightarrow y^{11}(-1)\times (-2)+32=32\times 31 y^{11}(-1)=\frac{30\times 32}{-2}=480 \\ \Rightarrow y^1(-1)-y^{11}(-1)=496 \end{gathered}$$