Let $f(x) = 1-x$. Then, the goal is to find $$g(x)=a_0 + a_1x+a_2x^2 + a_3x^3+\cdots$$

satisfying $f(x)g(x)=1$. $$\begin{align*} 1= f(x)g(x) =\ & (1-x)g(x) \\ =\ & g(x) - xg(x) \\=\ & (a_0 + a_1x+a_2x^2 + \cdots) - x(a_0 + a_1x+a_2x^2 + \cdots)\\ 1 =\ & a_0 + (a_1-a_0)x + (a_2-a_1)x^2 + (a_3-a_2)x^3 + \cdots \end{align*}$$

Therefore, it must hold that $$a_0 =1,\quad\text{and}\quad a_{i+1}-a_i=0,\ i=1,2,3,\cdots$$

This means that $a_0=a_1=a_2=a_3=\cdots = 1$ or $$g(x)=1+x+x^2+x^3+\cdots$$