Curves contained in a plane spanned by its derivatives Suppose $\boldsymbol{r}(t)$ is a smooth vector-valued function $I\rightarrow\mathbb{R}^n$, i.e. a curve. For each $t\in I$, $\frac{{\rm d}^k\boldsymbol{r}}{{\rm d}t^k}$ is a linear combination of $\frac{{\rm d}\boldsymbol{r}}{{\rm d}t},\cdots,\frac{{\rm d}^{k-1}\boldsymbol{r}}{{\rm d}t^{k-1}}$ and $\frac{{\rm d}\boldsymbol{r}}{{\rm d}t},\cdots,\frac{{\rm d}^{k-1}\boldsymbol{r}}{{\rm d}t^{k-1}}$ are linearly independent. Then the whole curve is contained in an invariant plane spanned by $\frac{{\rm d}\boldsymbol{r}}{{\rm d}t},\cdots,\frac{{\rm d}^{k-1}\boldsymbol{r}}{{\rm d}t^{k-1}}$.
My book says that it suffices to prove that the derivatives of the basis $\{\frac{{\rm d}\boldsymbol{r}}{{\rm d}t},\cdots,\frac{{\rm d}^{k-1}\boldsymbol{r}}{{\rm d}t^{k-1}}\}$ are linear combinations of the basis itself, which is obviously. But I don't know why it works. Thanks for any help in advance.

First, to be careful, it's an affine plane (not necessarily passing through the origin).
Note that the $(k-1)$-plane spanned by the first $k-1$ derivatives is constant as a function of $t$, and for any vector $A$ orthogonal to that plane, we have $A\cdot \big(r(t)-r(t_0)\big)=0$. Can you finish?
