Statement from a textbook:
Let $n \in \mathbb N$ and let $a,b$ be extended real numbers with $a<b$. If $f:(a,b)\rightarrow \mathbb{R},$ and if $f^{(n+1)}$ exists on $(a,b)$, then for each pair of points $x,x_0\in(a,b)$ there is a number $c$ between $x$ and $x_0$ such that
$$f(x)=f(x_0)+\sum_{k=0}^n\frac{f^{(k)}(x_0)}{k!}(x-x_0)^k+\frac{f^{(n+1)}(c)}{(n+1)!}(x-x_0)^{(n+1)}.$$
My proof:
Suppose $x_0<x\in(a,b)=I$. Since $f^{(n+1)}$ exists on $I$, $f\in C^j(I)$ for each $j\in \{0,1,\ldots,n\}$. In particular $f^{(j)}$ is continuous on $[x_0,x]$ and differentiable on $(x_0,x)$. There exists $\alpha \in \mathbb{R}$ such that $G^{(j)}(u):=\sum_{k=j}^nf^{(k)}(x_0)\Delta u^{k-j}/(k-j)! + \alpha\Delta u^{(n+1-j)}/(n+1-j)!$ satisfies $G^{(0)}(x)=f(x)$ where $u\in [x_0,x]$ and $\Delta u = u-x_0$. Notice $G\in C^j(I)$ for each $j\in\{0,1,\ldots,n\}$, $G^{(j)}(x_0)=f^{(j)}(x_0)$, and $G^{(n+1)}(u)=\alpha$.
Define the function $H:[x_0,x]\rightarrow \mathbb{R}$ by $H(u):=(G-f)(u)$. Since $H(x_0)=H(x)$, $H\in C^0[x_0,x]$, and $H$ is differentiable on $(x_0,x)$, by Rolle's Theorem there exists a $c_1\in(x_0,x)$ such that $H^{(1)}(c_1)=0$. Since $H^{(1)}\in C^0[x_0,c_1]$ and differentiable on $(x_0,c_1)$, $H^{(2)}(c_2)=0$ for some $c_2\in(x_0,c_1)$. This follows from the definition of $G^{(1)}$ at $x_0$ and Rolle's Theorem.
Continuing in this same manner, the process stops when finally a $c\in(x_0,c_n)\subset(x_0,x)$ is chosen that satisfies $H^{(n+1)}(c)=0$, and hence $\alpha = G^{(n+1)}(c)=f^{(n+1)}(c).$
My main questions are if moving from $c_2$ to $c_n$ is valid and if my definitions of $G$ and $H$ satisfy the conditions required for Rolle's Theorem.
Updated proof:
Define the function $G:(a,b)\rightarrow \mathbb{R}$ by $G(u):=f(x_0)+\sum_{k=1}^n\dfrac{f^{(k)}(x_0)}{k!}(u-x_0)^k+ \dfrac{\alpha (u-x_0)^{n+1}}{(n+1)!}$ where $\alpha \in \mathbb{R}$ is the unique real number that satisfies $G(x)-f(x)=0$.
Clearly $G^{(j)}(x_0)-f^{(j)}(x_0)=0$ for all $j\in \{0,1,\ldots,n\}$. In particular $G(x_0)-f(x_0)=0$, and hence by either Rolle's Theorem or Generalized Mean Value Theorem (steps not shown) there exists a $c_1\in(x_0,x)$ such that $G^{(1)}(c_1)-f^{(1)}(c_1)=0$. Repeating $n-1$ more times will show $G^{(n+1)}(c)-f^{(n+1)}(c)=0$ for some $c\in(x_0,c_{n-1})$. Since $G^{(n+1)}(u)=\alpha$, the theorem is proven.