Just a thought: In the Taylor expansion of an analytic function $f(x)$, the $\Gamma(n+1) = n!$ appears in the coefficient for $x^n$. So if we use a Puiseux series instead, would we get a $\Gamma(n/k)$ appearing in the coefficient for $x^{n/k}$?
I searched around and the only related result is that the Gamma function values appear in the general Newton binomial expansion theorem.
Best Answer
If you take a function $f(z)$ with a converging Puiseux series in powers of $z^{1/k}$, say
$$ f(z) = \sum_{j=0}^\infty a_j z^{j/k}$$ then $$ f(z^k) = \sum_{j=0}^\infty a_j z^j$$ is analytic in a neighbourhood of $0$, with $$ a_n = \frac{1}{n!} \left. \dfrac{d^n}{dz^n} f(z^k) \right|_{z=0} $$
No, there's no reason for this to involve $\Gamma(n/k)$.