Let $f:U\to \Bbb R^p$ be a smooth function where $U$ is an open subset of $\Bbb R^n$. For $k \in \Bbb N$, let $$C_k=\{x\in U: \textrm{all partial derivatives of}~f~\textrm{of order}~\leq k~ \textrm{vanish at}~x\}.$$
Let $I^n$ be a cube in $U$. For $x\in C^k \cap I^n$ and $h\in \Bbb R^n$ with $x+h \in I^n$, how can we show that $$ |R(x,h)|\leq c\cdot |h|^{k+1}$$ for some constant $c$ only depending on $f$ and $I^n$, where $R(x,h)=f(x+h)-f(x)$, using Taylor's Theorem and compactness of $I^n$?
This question arises in Milnor's Topology from the Differentiable Viewpoint, p.19(Sard's theorem). Thanks in advance.
Taylor expansion gives \begin{equation}\begin{aligned} R(x, h)&=\sum_{\ell=1}^\infty \frac1{\ell!}\sum_{j_1=1}^n\dots\sum_{j_\ell=1}^n \prod_{i=1}^\ell \left(h_{j_i}\frac{\partial}{\partial x_{j_i}}\right) f(x)\\ &=\sum_{\ell=1}^k\frac1{\ell!}\sum_{j_1=1}^n\dots\sum_{j_\ell=1}^n \prod_{i=1}^\ell \left(h_{j_i}\frac{\partial}{\partial x_{j_i}}\right) f(x)\\ &+ \frac1{k!}\int_{x}^{x+h}dy_{j_1}\prod_{i=2}^{k+1}(y-x)_{j_i}\left(\prod_{i=1}^{k+1}\partial_{j_i}\right) f \end{aligned}\end{equation} The second line follows from integration by parts.
So, since $x\in C_k$, we know \begin{equation}\begin{aligned} R(x, h)&=\frac1{k!}\int_Idy_{j_1}\prod_{i=2}^{k+1}(y-x)_{j_i}\left(\prod_{i=1}^{k+1}\partial_{j_i}\right) f\\ &\leqslant |\!|h|\!|^{k+1}\frac{1}{k!}\sup_{x\in I,\vec j\in\{1,\dots,n\}^{k+1}}\partial^\ell_{\vec j}f. \end{aligned}\end{equation} The supremum over the derivatives of $f$ is finite because $I$ is compact.