I am trying to solve this  [Komal problem 661]:

>Let $K$ be a fixed positive integer. Let $(a_{0},a_{1},\cdots )$ be the sequence of real numbers that satisfies $a_{0}=-1$ and
$$\sum_{i_{0},i_{1},\cdots,i_{K}\ge 0,i_{0}+i_{1}+\cdots+i_{K}=n}\dfrac{a_{i_{1}}a_{i_{2}}\cdots a_{i_{K}}}{i_{0}+1}=0$$
for every postive integer $n$. Show that $a_{n}>0$ for $n\ge 1$.


**Add edit**:For the Iosif Pinelis point out,$b_{0}=(-1)^K$,Now I have known where my wrong,and Ira Gessel point that, Now How to prove 

Let $\displaystyle f(x)\triangleq\sum_{i\geq 0} a_i x^i$ and $\displaystyle g(x)\triangleq \sum_{i\geq 0} \cfrac{x^{i}}{i+1}$. 

Then, we get 
$$ f(x)^Kg(x) = \sum_{n\geq 0}b_nx^n \text{ with } b_n=\sum_{\substack{i_{0},i_{1},\cdots,i_{K}\ge 0\\i_{0}+i_{1}+\cdots+i_{K}=n}}\dfrac{a_{i_{1}}a_{i_{2}}\cdots a_{i_{K}}}{i_{0}+1}.$$
Since $b_n=0$ for $n\geq 1$, we get
\begin{align} 
f(x)^Kg(x)&=b_0=(-1)^K\\
\implies \left(\sum_{i\geq 0} a_i x^i\right)^K&=\frac{(-1)^K}{g(x)}=\frac{-(-1)^K\cdot x}{\sum_{i\geq 1} -\cfrac{x^{i}}{i}}=\frac{-(-1)^K\cdot x}{-\ln(1-x)}\\
\implies\sum_{i\geq 1} a_i x^i &=-a_{0}+ \left(\frac{-(-1)^Kx}{\ln(1-x)}\right)^{1/K}
=1-\left(\dfrac{-x}{\ln{(1-x)}}\right)^{1/K}\end{align}
Thus, using the Tyalor series expansion,





$$\dfrac{d^i}{dx^i}\left(1-\left(-\dfrac{x}{\ln{(1-x)}}\right)^{\frac{1}{K}}\right)|_{x=0}>0\tag{1}$$

But the last maybe it not easy prove it,can help me to prove $(1)$?Thanks


[Komal problem 661]:https://www.komal.hu/feladat?a=feladat&f=A661&l=en