Taylor series for tetration

$x \uparrow \uparrow m$ is never analytic at $x = 0$ for $m > 1$. For example, consider $m = 2$, then you have the function $x \uparrow \uparrow 2 = x^x = e^{x \ln x}$ which clearly has a branch point at $x = 0$ and hence is not analytic. You can see this if we substitute $\ln x = 2\pi i n + \log x$ where $\log$ represents the principal branch of the logarithm, in which case we have $x^x = e^{x (2\pi i n + \log x)} = e^{x 2 \pi i n} e^{x\log x}$, which certainly depends on $n$ as long as $x$ is not an integer. You have similar problems for all $m > 1$.

The function $x \uparrow \uparrow m$ is always analytic at $1$, which can be seen inductively: Clearly for $m=1$ it is analytic at $1$. If $x \uparrow \uparrow m$ is analytic at 1, then $x \uparrow \uparrow{(m+1)} = e^{(\ln x)(x\uparrow \uparrow m)}$ is analytic at $1$ because $\exp$ is analytic everwhere and $\ln$ is analytic at 1. The Taylor series of $x \uparrow \uparrow m$ at $1$ is not very nice, but the terms can be computed explicitly. Wolfram Mathworld tells us that $$ (e^x) \uparrow\uparrow m = \sum_{n=0}^m \frac{(n+1)^n}{(n+1)!} x^n + \sum_{n = m+1}^\infty a_{m n} x^n$$ where $a_{m n}$ have a rather complicated recursive formula. To compute the Taylor series of $x \uparrow\uparrow m$ at $1$, you could compose the above series with the series for $\ln(x)$ at $1$ using Faa di Bruno's formula. Not very simple, but it probably as close to a closed form as you will get.