Question 52
Let $f(x) = x^{2} + 2x + 1$. Let $g(x) = f(f(\cdots f(x)))$, where there are $2009$ $f$'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 $a_{2^{2009}-1}$.
$$
g(x) = x^{2^{2009}} + a_{2^{2009}-1} x^{2^{2009}-1} + \cdots + a_{1} x + a_{0}
$$
where the $a_{i}$ are constants. Compute $a_{2^{2009}-1}$.