How do I estimate log10 of log10 of 8 billionth element of A000670? How do I estimate log10 of log10 of the 8 billionth element of A000670 
as an actual number? I believe it's between 10 and 100, but am not 
sure why I believe this. 
I realize http://oeis.org/A000670 offers some approximations of 
A000670 (in terms of n), but nothing that quickly leads to an actual 
value of log10(log10(a(n))) 
Goal: I'm trying to calculate the number of possible "prejudices" in 
the world, and it's the nth element of A000670, where n is the current 
world population. I don't need an exact number, just something like 
10^(10^r) [to the nearest integer value for r], and I'm estimating n at about 8 billion right now. 
Ideally, I'd like a good general estimate of log10(log10(a(n))) but 
that may be asking too much. 
EDIT: It's easy to show A000670 is bounded by 2^n*n! (aka http://oeis.org/A000165), but this bound is loose and unhelpful.
http://oeis.org/A000670 notes "Unreduced denominators in convergent to log(2) = lim[n->inf, na(n-1)/a(n)].". Does this mean that: n!*log(2)^n would be an estimate, at least in terms of order of magnitude? It seems a bit low. Since log(2) < 1 this can't be right. Maybe 
n!/(log(2)^n)

 A: See for instance Wilf's generatingfunctionology, p. 175, Example 5.2.1 "Ordered Bell Numbers" for the asymptotic
$$a(n) = n! \left(\frac{1}{2 \ln(2)^{n+1}} + O(0.16^n)\right).$$
It just arises from looking at the "leading singularity" of the exponential generating function. This approximation is particularly excellent since the first correction grows exponentially while the remaining correction decays exponentially.
Heuristically, Stirling's approximation says $\log_{10} \log_{10} (8 \cdot 10^9)!$ is going to be on the order of the exponent, namely 9--in fact, Mathematica tells me it's 10.879...--and the corrections are only going to contribute perhaps a few percent since even exponential growth is nothing compared to the growth of $n!$, so the nearest integer is definitely going to be 11.
We can be more precise though and actually prove it. Singularity analysis more fully gives
$$a(n) = \frac{n!}{2} \sum_{k=-\infty}^\infty \frac{1}{(\ln 2 + 2 \pi i k)^{n+1}},$$
the $k=0$ term of which gives the asymptotic above. Hence
$$\frac{1}{n!}\left|a(n) - \frac{1}{2 \ln(2)^{n+1}}\right| \leq \sum_{k=1}^\infty |\ln 2 + 2\pi i k|^{-(n+1)} < \sum_{k=1}^\infty (2\pi k)^{-(n+1)} = \frac{\zeta(n+1)}{(2\pi)^{n+1}}$$
where $\zeta$ is the Riemann zeta function. This says
$$\log_{10} \log_{10} a(8 \cdot 10^9) = \log_{10} \log_{10} (8 \cdot 10^9)! \left(\frac{1}{2 \ln(2)^{8 \cdot 10^9+1}} \pm \frac{\zeta(8 \cdot 10^9 + 1)}{(2\pi)^{8 \cdot 10^9}}\right).$$
Mathematica is happy to compute $\log_{10} \log_{10}$ of this expression, say with 40 digits of precision. I've included some related expressions too.

So, yup, the answer is 11.
A: How close do you want to get?  If we use Stirling on your upper bound, we want $\log_{10}\log_{10}\frac{2^{8E9}8E9^{8E9}}{e^{8E9}}\sqrt {2\pi 8E9}= \log_{10}8E9(\frac 12+ \log_{10}\frac {16E9}e)+\frac 12\log_{10} 2\pi\approx 9+\log_{10}8+1\approx 11$
