Let $m \ge 0$ be a positive integer, and let $0 < x < r$ be real numbers. Then, we have the identity $$\frac{r}{r-x} = \sum_{n=0}^\infty x^nr^{-n}$$ for all $x \in (-r, r)$. Using proposition 4.2.6, conclude the identity
$$\frac{r}{(r-x)^{m+1}} = \sum_{n=m}^\infty \frac{n!}{m!(n-m)!}x^{n-m}r^{-n}$$ for all integers $m \ge 0$ and $x \in (-r, r)$. Also explain why the series on the right-hand side is absolutely convergent.
Proposition 4.2.6 says that a real analytic function is $k$ times differentiable and the $k$th derivative is given by $$f^{(k)}(x) = \sum_{n=0}^\infty c_{n+k} \frac{(n+k)!}{n!}(x-a)^n,$$ if $f(x) = \sum_{n=0}^\infty c_n (x-a)^n$.
I can show that $(r-x)^m = \sum_{k=0}^m \frac{m!}{k!(m-k)!}(-x)^{m-k} r^k$, using Binomial formula. Then, $$\frac{r}{(r-x)^{m+1}} = \frac{1}{\sum_{k=0}^m \frac{m!}{k!(m-k)!}(-x)^{m-k} r^k}\sum_{n=0}^\infty x^nr^{-n} .$$ But, I am not sure how the right hand side is equal to $\sum_{n=m}^\infty \frac{n!}{m!(n-m)!}x^{n-m}r^{-n}$. I also don't know how to use Proposition 4.2.6.
I appreciate if you give some help.
Let $g(x)=\frac{r}{r-x}$. Then $g'(x)=\frac{r}{(r-x)^2}$, $g''(x)=\frac{2r}{(r-x)^3}$, $g^{(3)}(x)=\frac{6r}{(r-x)^4}$, and in general you should be able to prove by induction on $m$ that $g^{(m)}(x)=\frac{m!r}{(r-x)^{m+1}}$.
On the other hand, you can apply Proposition 4.2.6 to $g(x)$ with $k=m$, $c_n=r^{-n}$ for each $n$, and $a=0$ to get
$$g^{(m)}(x)=\sum_{n\ge 0}r^{-(n+m)}\frac{(n+m)!}{n!}x^n\;,$$
and it’s straightforward to show that this is equal to
$$\sum_{n\ge m}r^{-n}\frac{n!}{(n-m)!}x^{n-m}\;.$$
Now just equate the two expressions for $g^{(m)}(x)$.