Why is $\exp(n/2 * (\log 2)^2) \approx 1.27^n$

Question

I saw this approximation made in https://stats.stackexchange.com/questions/473496/infinite-coin-toss-probability by the accepted answer.

What inspired this approximation?

Answer 1

Recall that $\ln x^a = a \ln x$ and $x^a = e^{a \ln x}.$

So, we can rewrite this expression as $e^{n \ln\left(2^{\frac{\ln 2}{2}}\right)} = \left(2^{\frac{\ln 2}{2}}\right)^n \approx 1.27^n.$