Is my derivation of Taylor's Formula in $\mathbb R$ for the remainder acceptable?

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 $(x_0,x)$ for all $x_0<u\le x$. For each $j\in \{0,1,\ldots,n\}$ set $m_j:=f^{(j)}(x_0)$. Choose $\alpha \in \mathbb{R}$ such that $G^{(j)}(u):=\sum_{k=j}^nm_k\Delta u^{k-j}/(k-j)! + \alpha\Delta u^{(n+1-j)}/(n+1-j)!$ satisfies $G^{(0)}(x)=f(x)$, where $\Delta u = u-x_0$. Notice $G^{(j)}(x_0)=m_j$, $G^{(n+1)}(u)=\alpha$, and $G^{(j)}\in C^j(I)$ for each $j\in\{0,1,\ldots,n\}$ and $u\in [x_0,x]$.

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 $H^{(1)}\in C^1$ 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 $H^{(n+1)}(c)=0$. But this implies $\alpha = G^{(n+1)}(c)=f^{(n+1)}(c).$

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.

Note: I am a structural engineer with no formal math credentials. I consider myself a math wanna-be and would like to move past this problem to reach the chapter on integration. I have been working on this problem for about eight weeks without any assistance, and this is what I have come up with that made the most rational sense to me.