when I 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 $n\ge 1$

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\\ \implies \left(\sum_{i\geq 0} a_i x^i\right)^K&=\frac{-1}{g(x)}=\frac{x}{\sum_{i\geq 1} -\cfrac{x^{i}}{i}}=\frac{x}{\ln(1-x)}\\ \implies\sum_{i\geq 0} a_i x^i &= \left(\frac{x}{\ln(1-x)}\right)^{1/K}. \end{align} Thus, using the Tyalor series expansion, $$ a_i=\frac{1}{i!}\frac{d^i}{dx^i}\left(\frac{x}{\ln(1-x)}\right)^{1/K} \Bigg|_{x=0}>0\iff \frac{d^i}{dx^i}\left(\frac{x}{\ln(1-x)}\right)^{1/K} \Bigg|_{x=0}>0.$$

where this wrong?