Theorem: Let $k\in\mathbb{N}$, $x_{0},x\in A\subseteq\mathbb{R}^{n}$, $x_{0}\neq x$ and $f:A\to\mathbb{R}$ satisfy that $\partial^{\alpha}f$ exists and is differentiable on $L=\{(1-t)x_{0}+tx\in\mathbb{R}^{n}\mid t\in[0,1]\}$ for $\lvert\alpha\rvert\leq k$. Then there exists $y\in\{(1-t)x_{0}+tx\in\mathbb{R}^{n}\mid t\in(0,1)\}$ such that:
\begin{equation}
f(x)=\hspace{-3pt}\sum_{\lvert\alpha\rvert\leq k}\frac{\partial^{\alpha}f(x_{0})}{\alpha!}(x-x_{0})^{\alpha}\hspace{1pt}+\hspace{-6pt}\sum_{\lvert\alpha\rvert=k+1}\hspace{-6pt}\frac{\partial^{\alpha}f(y)}{\alpha!}(x-x_{0})^{\alpha}.
\end{equation}
Proof: From the assumptions it follows that $\partial^{\alpha}f$ differentiable on $L$ for $\lvert\alpha\rvert\leq k$, and thus $L\subseteq A$ og $x_{0},x\in A$ are interior points. Moreover, there exists $r>0$ such that $F:(-r,1+r)\to\mathbb{R}$ is well defined by $F(t)=f\left((1-t)x_{0}+tx\right)$ and $k+1$-times differentiable on $[0,1]$.
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My question: What ensures that $F$ is $k+1$-times differentiable on $[0,1]$???
You write that $\partial^\alpha f$ exists and is differentiable on $L$ for $|\alpha| \leq k$. Does that not mean that $f$ is differentiable $k$ times, and then THAT derivative is differentiable once more on $L$? Since $f$ is then differentiable $k+1$ times, $F$ is too.
Sorry if I misunderstood in my answer, I cannot leave comments yet as my reputation is too low.