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

$$ \lim_{K \to \infty} h(x) = \left( \dfrac{x-1}{(1-1/d) x - 1}\right)^d $$