Formal Powerseries: Compute the reciprocal of $1-A(z)$ for $a_0 \ne 0$ 
Let $A(z)$ be a formal powerseries with $a_0=0$. Show that the reciprocal of $1-A(z)$ is given by
$$B(z) := \sum_{n=0}^\infty A(z)^n =  1 + A(z)^1 + A(z)^2 + \ldots$$

I realise that when we rewrite $C(z) := 1-A(z)$, then the reciprocal of $C$ has to exist since $c_0 \ne 0$. If we let $D(z) := C(z)B(z)$ we are done if we can show that $d_0 = 1$ and $d_n = 0$ for all $n > 0$. We can readily see that $b_0, c_0 = 1$, so $d_0 = 1$. So we are left to compute the remaining $d_n$ for $n > 0$ via the formula
$$d_n = \sum_{k=0}^n c_k b_{n-k}.$$
It is clear that $c_0 = 1$ and $c_n = -a_n$ for $n > 1$, but I do not see how to find a formula for the $b_n$ with $n > 1$. Could you please help me?

The formula for $b_n$ is a little unwieldy, but it is possible to prove this without computing $b_n$ exactly.  We use the same proof idea for the corresponding statement for real numbers:
$$
\frac1{1-x}=1+x+x^2+\dots\quad (|x|<1)
$$
The proof is
$$
\begin{align}
(1-x)(1+x+x^2+\dots)
&=(1-x)\left(\lim_{N\to\infty}1+x+\dots+x^{N-1}\right)
\\&=\lim_{N\to\infty} (1-x)(1+x+\dots+x^{N-1})
\\&=\lim_{N\to\infty}(1-x^N)\tag{cancellation}
\\&=1
\end{align}
$$
We do this same proof, replacing $x$ with $A(z)$. However, in what sense can we say that $\lim_{N\to\infty} [A(z)]^N=0$? These are just formula power series, they do not necessarily have real number values.
Still, the concept of a limit makes sense for formal power series. If $F_1(x), F_2(x),\dots$ is a sequence of power series, we say that $\lim_{n\to\infty} F_n(x)=G(x)$ if the following holds. Here, $F_n(x)=\sum_i f_{n,i}x^i$, and $G(x)=\sum_i g_ix^i$.

For all $i\in \mathbb N$, there exists $N(i)\in \mathbb N$ so $j\ge N(i)$ implies $f_{i,j}=g_i$.

This is the definition of limit used to define the infinite sum $\sum_{n=0}^\infty A(z)^n$ in the first place. Now, you should be able to show $A(z)^N\to 0$ in this sense. Here is where we need the fact that $a_0=0$; this ensures that $z^N$ divides $[A(z)]^N$, so that the first $N$ coefficients of $[A(z)]^N$ are zero.
You also need to show that you can bring the $1-A(z)$ inside the limit. That is, you need to use the general fact that $H(x)\lim_n F_n(x)=\lim_n H(x)F_n(x)$. This is straightforward to prove.
