(a) Let $g(x) = \sum_{n=0}^\infty \binom{k}{n} x^n$. Differentiate this series to show that $$g'(x) = \frac{kg(x)}{1+x}, \quad -1 < x < 1.$$ (b) Let $h(x) = (1+x)^{-k} g(x)$ and show that $h'(x) = 0$.
(c) Deduce that $g(x) = (1+x)^k$.
For part (a), I'm getting $g'(x) = g(x) \frac{n}{x}$. How did they get $g'(x) = \frac{kg(x)}{x+1}$?
\begin{align*} g(x) &= \sum_{n=0}^\infty \binom{k}{n} x^n \\ \implies g'(x) &= \sum_{n=1}^\infty n \binom{k}{n} x^{n-1}. \end{align*} Note that, as you've done, you can't pull $n$ out the sum here. $n$ is a dummy variable; $g'(x)$ cannot depend on $n$. Instead, note \begin{align*} (1+x) g'(x) &= \sum_{n=1}^\infty \left(n \binom{k}{n} x^{n-1} + n \binom{k}{n} x^n\right) \\ &= \sum_{n=1}^\infty \left(\frac{n k!}{(k-n)! n!} x^{n-1} + \frac{nk!}{(k-n)! n!} x^n\right) \\ &= k \sum_{n=1}^\infty \left(\binom{k-1}{n-1} x^{n-1} + \binom{k-1}{n-1} x^n\right) \\ &= k \sum_{n=0}^\infty \left( \binom{k-1}{n} x^n + \binom{k-1}{n-1} x^n\right) \end{align*} by reindexing the first term. But by Pascal's rule, we obtain $$(1+x)g'(x) = k \sum_{n=0}^\infty \binom kn x^n = kg(x).$$