If $$f(x)-f(x_0)=g(x)(x-x_0)$$ and $g \in C^{(n-1)}(U(x_0))$, where $U(x_0)$ is a neighbourhood of $x_0$, then show that $f(x)$ has a derivative $f^{(n)}(x_0)$ of order $n$ at $x_0$.
I think since $f(x)$ is continuous on its derivative of order $n-1$, I think I just need to verify that $f^{(n-1)}(x)-f^{(n-1)}(x_0)=g^{(n-1)}(x)(x-x_0)$ works and completes the proof. How do I continue it? Thanks in advance.
Hint: Note that \begin{align} f(x) = f(x_0)+g(x)(x-x_0) \ \ \implies \ \ f^{(n-1)}(x) = g^{(n-1)}(x)(x-x_0)+(n-1)g^{(n-2)}(x) \end{align} which means \begin{align} \frac{f^{(n-1)}(x)-f^{(n-1)}(x_0)}{x-x_0} = g^{(n-1)}(x)+(n-1)\frac{g^{(n-2)}(x)-g^{(n-2)}(x_0)}{x-x_0}. \end{align}