Proof: By Induction

By hypothesis, $I\subset\mathbb R$ is an interval and $f:I\to\mathbb R$ is a $(n+1)$ times continuously differentiable function where $n\ge 0$ is a natural number.
Let $a,x\in I,$ and $n!$ denote the factorial, and $f^{\{n\}}$ denote the $n$-th derivative. We prove the statement by induction on $n.$
Base case $n=0$
- The fundamental theorem of calculus gives us $$f(x)=f(a)+\int_a^x f^{\prime}(t)dt.$$
Induction step $n\to n+1$
- By the induction hypothesis and¹ by integration by parts $$\begin{align} R_n(x)&=\frac 1{(n-1)!}\int_a^x (x-t)^{n-1}f^{\{n\}}(t)dt\nonumber\\ &=-\int_a^xf^{\{n\}}\frac{d}{dt}\left(\frac{(x-t)^n}{n!}\right)dt\nonumber\\ &=-f^{\{n\}}(t)\frac{(x-t)^n}{n !}\;\Rule{1px}{4ex}{2ex}^{t=b}_{t=a}+\int_a^x \frac{(x-t)^n}{n !}df^{\{n\}}(t)\nonumber\\ &=\frac{f^{\{n\}}(a)}{n !}(x-a)^n+\frac 1{n !}\int_a^x (x-t)^{n}f^{\{n+1\}}(t)dt\nonumber\\ &=R_{n+1}(x)\nonumber \end{align}$$
It follows for all $n\ge 0$ that $$f(x)=f(a)+\sum_{k=1}^n \frac{f^{\{k\}}(a)}{k!}(x-a)^k+R_{n+1}(x).$$

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In the mnemonic notation of the partial integration, $\int fdg=fg-\int gdf$ replace $f$ by $f^{\{n\}}$ and $g$ by $\frac{(x-t)^n}{n!}.$ ↩