The first few terms of the Maclaurin series can be obtained explicitly:
$$1-x-{\frac { \left( d-1 \right) \left( K-1 \right) }{2\,Kd+2}}{x}^{2 }-{\frac { \left( d-1 \right) \left( d-2 \right) \left( K-1 \right) \left( K-2 \right) }{ \left( 6\,Kd+6 \right) \left( Kd+2 \right) }}{ x}^{3}+O \left( {x}^{4} \right) $$
The coefficient of $x^2$ is not especially small, but of course on a small interval around $0$ the $1-x$ dominates. I guess the question is really about the behaviour as $K \to \infty$ for fixed $x$.
The Maclaurin coefficient of $x^n$ in $h(x)$ is
$$ A_n(K,d) = \sum_{ k=0}^n (-1)^k {K \choose k} \dfrac{\Gamma(K+n-k) \Gamma(Kd-d+n-k) \Gamma(Kd)}{\Gamma(K) \Gamma(Kd-d) \Gamma(Kd+n-k) (n-k)!}$$
$A_n(K,d)$ is a rational function of $K$ and $d$ where it appears that numerator and denominator both have degree $n-1$ in $K$ for $n \ge 1$. Thus as $K \to +\infty$, $h(x)$ may be approaching a smooth limit. In fact, it looks to me like
$$ \lim_{K \to \infty} h(x) = \left( \dfrac{x-1}{(1-1/d) x - 1}\right)^d $$