Proof

By hypothesis, $I\subset\mathbb R$ is an interval and $f:I\to\mathbb R$ is a $n$ times continuously differentiable function .
Let $a\in I.$
The Taylor's formula with remainder term of lagrange of the $n$-th order yields $$\begin{align} f(x)-\sum_{k=0}^{n-1} \frac{f^{\{k\}}(a)}{k !}(x-a)^k&=\frac{f^{\{n\}}(\xi)}{n !}(x-a)^{n}\nonumber\\ &=\frac{f^{\{n\}}(a)}{n !}(x-a)^{n}+\frac{f^{\{n\}}(\xi)-f^{\{n\}}(a)}{n !}(x-a)^{n}\nonumber.\end{align}$$
Define a function $r:I\to\mathbb R$ (note: the value of $\xi$ depends on every $x$, since it lies between $a$ an $x$): $$r(x):=\frac{f^{\{n\}}(\xi)-f^{\{n\}}(a)}{n !}.$$
It follows from the hypothesis that the $n$-th derivative $f^{\{n\}}$ is continuous.
Because of the continuity of functions is preserved for arithmetic operations, and because $\xi$ lies between $a$ an $x,$ we get for the limit $$\lim_{x\to a} r(x)=\lim_{\xi\to a} \frac{f^{\{n\}}(\xi)-f^{\{n\}}(a)}{n !}=0.$$
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Forster Otto: "Analysis 1, Differential- und Integralrechnung einer Veränderlichen", Vieweg Studium, 1983