Is taking a Taylor expansion of $x^{x^x}$ possible? [closed]

Question

Disclaimer: I do not know that much math, so there might be obvious mistakes that I make in this question.

As far as I can tell, taking the derivative of $x^{x^x}$ infinitely is possible. However, this does not mean a Taylor series is possible. So what I am asking is, is it possible. I see 5 possible outcomes:

1. It is possible, and the answer is known/easy enough to figure out
2. It has been proven to be possible, but we don’t know what it is yet
3. It has been proven to be impossible, and cannot be done
4. It may or not be possible, we don’t know yet
5. Other

So, which one of these is it?

Answer 1

It is possible and it is the same for all tetrations. In this case, around $x=0$, it would be $$x^{x^x}=x+\sum_{n=2}^p \frac {t^n}{(n-1)!} P_n(t)\, x^{n}+O(x^{p+1})\qquad \text{with} \qquad t=\log(x)$$ and the first polynomials are $$\left( \begin{array}{cc} 2 & 1 \\ 3 & t+1 \\ 4 & t^2+3 t+1 \\ 5 & t^3+6 t^2+7 t+1 \\ 6 & t^4+10 t^3+25 t^2+15 t+1 \\ 7 & t^5+15 t^4+65 t^3+90 t^2+31 t+1 \\ 8 & t^6+21 t^5+140 t^4+350 t^3+301 t^2+63 t+1 \end{array} \right)$$ where you can notice interesting patterns.

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