HMMT November 2009 U21

Let $f(x)=x^{2}+2 x+1$. Let $g(x)=f(f(\cdots f(x)))$, where there are $2009 f \mathrm{~s}$ in the expression for $g(x)$. Then $g(x)$ can be written as

$$g(x)=x^{2^{2009}}+a_{2^{2009}-1} x^{2^{2009}-1}+\cdots+a_{1} x+a_{0}$$

where the $a_{i}$ are constants. Compute $a_{2^{2009}-1}$.