Understanding power series multiplication From Enumerative Combinators by Stanley, one of his beginning examples begins is as follows:

Let $(\sum_n_\ge_0\alpha^nx^n)(1-\alpha x)=\sum_n_\ge_0c_nx^n$ where $\alpha$ is a non-zero complex number. Then by the definition of power series multiplaction,
  $c_0$ = 1 and $c_n=\alpha^n-\alpha(\alpha^{n-1})$ for $n\ge1$. 

And the argument continues and I understand the rest, however I cannot come up with the same values for $c_n$ as he does. For example, for $n=0$ I come up with $1-\alpha x=c_0$ which would be fine if the right hand side was $n=1$ but that isn't the case. 
My question: Can someone please spell out his argument for obtaining the values of $c_n$ for me?
 A: Hint: If the two series $\sum a_n$ and $\sum b_n$ are absolutely convergent then the product series $\sum c_n=(\sum a_n)(\sum b_n)$ is defined by
$$c_n=\sum_{k=0}^na_kb_{n-k}.$$
Take $a_n=\alpha^n$ and $b_0=1, b_1=-\alpha$ and $b_n=0\forall n\geq2$ then $c_0=1$ and $c_n=a_nb_0+a_{n-1}b_1=\alpha^n-\alpha(\alpha^{n-1})=0\forall n\geq 1$.
A: You have to group powers of $x$; the typical way to reconcile multiple series is to re-index them, as follows:
$$\begin{align}\left(\sum_{n \geq 0} \alpha^n x^n\right)(1 - \alpha x)
  &= \sum_{n \geq 0} \alpha^n x^n - \sum_{n \geq 0} \alpha^{n + 1} x^{n + 1} \\
  &= \sum_{n \geq 0} \alpha^n x^n - \sum_{m \geq 1} \alpha^m x^m \\
  &= \alpha^0 x^0 + \sum_{n \geq 1} \alpha^n x^n - \sum_{m \geq 1} \alpha^m x^m \\
  &= 1\end{align}$$
The "reindexing" was the definition $m = n + 1$ after the first line.
A: The simple argument is this calculation:
$$\begin{align*}
\left(\sum_{n\ge 0}\alpha^n x^n\right)(1-\alpha x)&=\sum_{n\ge 0}\alpha^n x^n-\sum_{n\ge 0}\alpha^{n+1}x^{n+1}\\\\
&=\sum_{n\ge 0}\alpha^n x^n-\sum_{n\ge 1}\alpha^n x^n\\\\
&=\alpha^0\\\\
&=1
\end{align*}$$
However, your wording suggests that he may be looking at the following definition using the Cauchy product: if 
$$\left(\sum_{n\ge 0}a_nx^n\right)\left(\sum_{n\ge 0}b_nx^n\right)=\sum_{n\ge 0}c_nx^n\;,$$
then
$$c_n=\sum_{k=0}^na_kb_{n-k}\;.$$
In your case $b_0=1$, $b_1=-\alpha$, and $b_k=0$ for $k>1$, so 
$$c_n=\sum_{k=0}^n\alpha^kb_{n-k}=\alpha^{n-1}b_1+\alpha^nb_0=-\alpha^n+\alpha^n=0$$
if $n\ge 1$, and
$$c_0=\sum_{k=0}^0\alpha^kb_{0-k}=\alpha^0=1\;.$$
