Rudin is proving Abel's Theorem on page 174 of my print edition
He writes,
Let $s_n = c_0 + c_1 + ... + c_n$, $s_{-1} = 0$. Then
$\sum_{n=0}^m c_n x^n = \sum_{n=0}^m (s_n - s_{n-1})x^n = (1-x) \sum_{n=0}^{m-1} s_n x^n + s_m x^m$
How did he arrive at $\sum_{n=0}^m (s_n - s_{n-1})x^n = (1-x) \sum_{n=0}^{m-1} s_n x^n + s_m x^m$ ?
And also, is that $(1-x) \sum_{n=0}^{m-1} (s_n x^n + s_m x^m)$ or $(1-x) \sum_{n=0}^{m-1} (s_n x^n) + s_m x^m$ ?
Quite simply, $$\begin{align*} \sum_{n=0}^m (s_n - s_{n-1})x^n &= \sum_{n=0}^m s_n x^n - \sum_{n=0}^m s_{n-1} x^n \\ &= s_m x^m + \sum_{n=0}^{m-1} s_n x^n - \sum_{n=-1}^{m-1} s_n x^{n+1} \\ &= s_m x^m + \sum_{n=0}^{m-1} s_n (x^n - x^{n+1}) \\ &= s_m x^m + (1-x) \sum_{n=0}^{m-1} s_n x^n. \end{align*}$$
The first step is distributing $x^n$ over $(s_n - s_{n-1})$ and splitting the result into two sums. The second step peels off the last term of the first sum, and rewrites the second sum by shifting the index of summation by $1$. The third step observes $s_{-1} = 0$ and combines the two sums from the second step, which now share the same lower and upper bound of summation. The last step factors out $(1-x)$.