jmc

Multiply both sides by $x-1$; the right hand side collapses by the reverse of the difference of squares. $$\begin{array}{rl}(x-1)(x^{2^{2008}-1}-1)g(x)& =(x-1)(x+1)(x^2+1)(x^4+1)\cdots (x^{2^{2007}}+1)-(x-1)\\ & =(x^2-1)(x^2+1)(x^4+1)\cdots (x^{2^{2007}}+1)-(x-1)\\ & =\cdots \\ & =\left(x^{2^{2008}}-1\right)-(x-1)=x^{2^{2008}}-x\end{array}$$Substituting $x=2$, we have$$\left(2^{2^{2008}-1}-1\right)\cdot g(2)=2^{2^{2008}}-2=2\left(2^{2^{2008}-1}-1\right)$$Dividing both sides by $2^{2^{2008}-1}$, we find $g(2)=\boxed{2}$.