2009 HMMT November Guts Round Problem 21

Let f(x) = x^2 + 2x + 1. Let g(x) = f(f(\cdots f(x))), where there are 2009 fs 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_1x + a_0,

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

Answers are checked by value, so any equivalent form is accepted: 1/2, \frac{1}{2} and 0.5 all count as the same answer.

Full credit goes to HMMT for authoring these problems. This problem is from the November 2009 contest; the official solution is available on the HMMT archive. HMMT is not affiliated with or endorsing TopsOJ in any way.