The binomial series expansion that you used -
$$
\frac1{1-y}=\sum_{n=0}^{\infty} y^n
$$
is valid only when $|y|<1$. Otherwise, the series diverges. In your solution, you are substituting in $(x-1)$ for $y$, so your solution is valid for $|x-1|<1$, so your power series is in some sense centred at $1$. Indeed, your power series is not a power series in $x$, but a power series in $x-1$. Normally, we want to find a series in $x$ (centred at $0$), and this is what the textbook solution is doing. If we substitute in $\frac{x}{2}$ for $y$, then we get a series valid for $|x/2|<1$, or for $|x|<2$.
Note, however, that you can transform from one series to the other. The following is non-rigorous, but can be tidied up if you want to:
\begin{align}
\sum_{n=0}^\infty (x-1)^n &= \sum_{n=0}^\infty \sum_{k=0}^n {n \choose k} x^k(-1)^{n-k}\\
&=\sum_{k=0}^\infty x^k \sum_{n=k}^\infty {n\choose k} (-1)^{n-k}
\end{align}
At this point, the series $\sum_{n=k}^\infty {n\choose k} (-1)^{n-k}$ doesn't actually converge, but if you extend the notion of summation in a special way (called 'Abel summation', I think) then you can get it to come out as $2^{-n-1}$. Not at all rigorous: has anyone got a better way of doing this?
