A curve is contained in a hyperplane if and only if $\kappa_{n-1} = 0$ I'm trying to prove that a curve $\gamma$ in $\mathbb{R}^n$ is contained in a hyperplane if and only if $\kappa_{n-1} = 0$, where $$\kappa_{i} = \frac{\langle e_{i}',e_{i+1}\rangle}{\| \gamma' \|}$$ is the $i^{th}$ curvature function of $\gamma$ (here $e_{i}$ is the $i$th vector in the Frenet Frame). My strategy is to show that $\kappa_{n-1} = 0$ if and only if the unit normal vector of the osculating hyperplane and the distance of the osculating hyperplane from the origin are constant. However, I'm not sure how to proceed. I can see why this result is true in 3 dimensions but am struggling to reframe it in a way that is easy to deal with in $n$ dimensions.

$\newcommand{\Reals}{\mathbf{R}}\newcommand{\eps}{\varepsilon}\newcommand{\Neg}{\phantom{-}}\newcommand{\Basis}{\mathbf{e}}\newcommand{\Brak}[1]{\left\langle #1\right\rangle}$Let $\eps > 0$, and $\gamma:(-\eps, \eps) \to \Reals^{n}$ a curve of class $C^{n}$ in $\Reals^{n}$, parametrized by arc length (for simplicity, and without loss of generality), whose first $(n - 1)$ derivatives are linearly independent. Define the Frenet-Serret frame $(\Basis_{i}(s))_{i=1}^{n-1}$ by applying Gram-Schmidt to the ordered basis
$$
\bigl(\gamma'(s), \gamma''(s), \dots, \gamma^{(n-1)}(s)\bigr),
$$
let $\Basis_{n}(s)$ denote the unique unit field orthogonal to $\Basis_{i}(s)$ for each $i = 1, \dots, n-1$ and forming a positive orthonormal frame. (Orientation isn't important below, it merely fixes one of two choices.)
For $1 \leq i \leq n - 1$, define
$$
\kappa_{i}(s) = \Brak{\Basis_{i}'(s), \Basis_{i+1}(s)},
$$
so that
$$
\left[\begin{array}{@{}c@{}}
\Basis_{1}' \\
\Basis_{2}' \\
\Basis_{3}' \\
\vdots \\
\Basis_{n}' \\
\end{array}\right]
  = \left[\begin{array}{@{}ccccc@{}}
0 & \Neg \kappa_{1}(s) & & & \\
-\kappa_{1}(s) & 0 & \Neg \kappa_{2}(s) & & \\
0 & -\kappa_{2}(s) & 0 & \ddots & \\
  &  & \ddots & 0 & \Neg \kappa_{n-1}(s) \\
  &  &  & -\kappa_{n-1}(s) & 0 \\
\end{array}\right]
\left[\begin{array}{@{}c@{}}
\Basis_{1} \\
\Basis_{2} \\
\Basis_{3} \\
\vdots \\
\Basis_{n} \\
\end{array}\right].
\tag{1}
$$
In words, $\Basis_{i}'(s)$ lies in the plane spanned by $\Basis_{i-1}(s)$ and $\Basis_{i+1}(s)$.
If $\kappa_{n-1}(s) \equiv 0$, then $\Basis_{n}'(s) \equiv 0$, i.e., $\Basis_{n}(s) = \Basis_{n}$ is constant, and
$$
\Brak{\gamma(s) - \gamma(0), \Basis_{n}} \equiv 0.
$$
(The left-hand side vanishes at $0$, and its derivative vanishes identically by (1).) This means $\gamma$ lies in the hyperplane
$$
\Brak{\mathbf{x} - \gamma(0), \Basis_{n}} = 0.
$$
Conversely, if $\gamma$ lies in some hyperplane
$$
\Brak{\mathbf{x} - \gamma(0), \Basis} = 0,
$$
then every derivative $\gamma^{(i)}(s)$ is orthogonal to $\Basis$. Particularly, $\kappa_{n-1}(s) = \Brak{\Basis_{n-1}'(s), \Basis} \equiv 0$ by (1).
