Old chestnut: f(f(x))=exp(x)

I came across this while browsing, and was amazed to find how difficult the subject of "iterative roots" can be. Can someone comment on the following?
Near x=0, exp(x)=x+1, so f(x)=x+0.5
f(0) = 0.5
f(f(0)) = f(0.5) = exp(0) = 1
f(f(0.5)) = f(1) = exp(0.5) = sqrt(e)
f(f(1)) = f(sqrt(e)) = exp(1) = e

These 4 results define a cubic, and, continuing ad nauseam, with n results defining a polynomial a_0+a_1*x+.....a_(n-1)*x^(n-1)

As n grows to infinity, we will get a power series for f(x).
Is this a good way to get a power series expansion for f(x)? Does it converge in a useful manner?

Alternatively, define f(x) as sum a_n * x^n. f(f(x)) has a power series
sum b_n * x^n, where the b_n can be expressed in terms of the a_n,and also as the terms in the series for exp. Can the a_n be calculated in this way? Is this a useful method? Does it give the same series as the first method above?

Thanks in advance,
Mike.

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