Big, Bigger, Biggest!

Define the function $f(x)=\log(\log(x))$ . Since $f(x)$ is increasing for $x>1$ , we can infer that if $f(x_{2})>f(x_{1})$ , then $x_{2}>x_{1}$ .

Now,

$f(x!) \approx \log(\log(\sqrt{2\pi}({10^{100}}^{(10^{100}+\frac{1}{2})}e^{-10^{100}}))$

$=\log(\log(\sqrt{2\pi}10^{(10^{102}+50)}e^{-10^{100}}))$

$=\log(\log\sqrt{2\pi}+10^{102}+50-10^{100}\log e)$

$\approx \log(10^{102})$

$=102$

Note that $10^{102}$ is larger than $10^{100}\log e$ by orders of magnitude of at least $2$ , so our approximation is justified.

We do the same thing to $a_{2}, a_{3}, a_{4}, a_{5}, a_{6}, a_{7}$ and find that

$f(a_{1}) \approx 102$

$f(a_{2})= 10^{100} +2$

$f(a_{3})= 200$

$f(a_{4})= 10^{100}+100$

$f(a_{5})= 100+ 10^{100} +10^{10^{100}}+ \log (\log e) \approx 100+ 10^{100} +10^{10^{100}}$ . (As an exercise, prove this)

$f(a_{6})= 0$

$f(a_{7})= 10^{10^{100}}$

All of these are relatively easy to prove. The readers are invited to verify the results.

It is now obvious that $f(a_{6})<f(a_{1})<f(a_{3})<f(a_{2})<f(a_{4})<f(a_{7})<f(a_{5})$ , from which the conclusion $a_{6}<a_{1}<a_{3}<a_{2}<a_{4}<a_{7}<a_{5}$ follows.

Therefore, the desired answer is $\boxed{6132475}$ .