Deriving formula for curvature of a curve in $\mathbb{R}^n$ I am trying to prove that given a parametric function of a regular curve in $\mathbb{R}^n$ the curvature of the curve for each $\gamma(t)$ is given by the following expression :
$$k(t)=\frac{\sqrt{||\gamma’(t)||^2||\gamma’’(t)||^2-(\gamma’(t)\cdot\gamma’’(t))^2}}{||\gamma’(t)||^3}$$
I have tried to use an arc-length parametrization of the curve but it leads me nowhere.
 A: There's no need to use arclength parametrization.  We have $\gamma'(t)=s'(t)T(t)$, where $T(t)$ is the unit tangent vector and from the definition of the Frenet frame in $\mathbb{R}^n$ we also have $\gamma''(t)=s''(t)T(t) + s'(t)^2\kappa(t)P_1(t)$, where $P_1(t)$ is the first normal vector in the $n$-dimensional Frenet frame.
Since $\{T,P_1\}$ is an orthonormal set,  $\|\gamma'\|^2 = (s')^2$ and $\gamma'\cdot\gamma''=s's''$ and also $\|\gamma''\|^2=(s'')^2+(s')^4\kappa^2$ .  The quantity you have under your square root comes out to
\begin{equation}
(s')^2(s'')^2+(s')^6\kappa^2-(s')^2(s'')^2=(s')^6\kappa^2.
\end{equation}
The result you are looking for follows.
There are nice formulas for all higher curvatures.  You could check out Section 9.1 in Differential Geometry of Curves and Surfaces by Banchoff and Lovett, 2nd edition. (The first edition does not have Chapter 9, so you need 2nd ed. or higher.)
A: There is nothing truly insightful: just a lot of computations. Reparameterize $\alpha$ such that $\alpha\circ \Phi$ is arc-length parameterized.  This yields:
\begin{equation}
\langle \alpha'(\Phi(u))\Phi'(u), \alpha'(\Phi(u))\Phi'(u) \rangle=1
\label{1}
\end{equation}
\begin{equation}
\langle \alpha''(\Phi(u))\Phi'(u)^2+\alpha'(\Phi(u))\Phi''(u), \alpha'(\Phi(u))\Phi'(u) \rangle=0
\label{2}
\end{equation}
The first relation also implies the two following equations:
\begin{equation}
\Phi'(u)=\frac{1}{\sqrt{\langle\alpha'(\Phi(u)),\alpha'(\Phi(u))\rangle}}=\frac{1}{|\alpha'(\Phi(u)|}
\end{equation}
$$\Phi''(u)=-\frac{1}{2}\frac{1}{\langle \alpha'(\Phi(u)),\alpha'(\Phi(u))\rangle^{\frac{3}{2}
}}2\langle \alpha'(\Phi(u)),\alpha''(\Phi(u))\rangle\Phi'(u)=-\frac{\langle \alpha'(\Phi(u)),\alpha''(\Phi(u))\rangle}{|\alpha'(\Phi(u))|^4}$$
With these four equations, we are ready to prove the main result. First let us introduce the variable $\Gamma=\alpha''(\Phi(u))\Phi'(u)^2+\alpha'(\Phi(u))\Phi''(u)$:
$$ |k|^2=|(\alpha(\Phi(u)))''|^2=|\Gamma|^2=\langle\Gamma,\alpha''(\Phi(u))\rangle\Phi'(u)^2+\langle\Gamma,\alpha'(\Phi(u))\rangle\Phi''(u)$$
By the second equation we have $ \langle\Gamma,\alpha'(\Phi(u))\rangle\Phi''(u)=0$ (because it is a regular parametrization and we may multiply and divide by $\Phi'(u)$). Thus:
$$k^2=\langle\Gamma,\alpha''(\Phi(u))\rangle\Phi'(u)^2=$$
$$\langle\alpha''(\Phi(u)),\alpha''(\Phi(u))\rangle\Phi'(u)^4+\langle \alpha'(\Phi(u)),\alpha''(\Phi(u))\rangle\Phi''(u)\Phi'(u)^2=$$
$$\frac{|\alpha''(\Phi(u))|^2}{|\alpha'(\Phi(u))|^4}-\frac{\langle \alpha'(\Phi(u)),\alpha''(\Phi(u))\rangle^2}{|\alpha'(\Phi(u))|^6}$$
Taking the square root from both sides yields:
$$k=\frac{\sqrt{|\alpha''(\Phi(u))|^2|\alpha'(\Phi(u))|^2-\langle \alpha'(\Phi(u)),\alpha''(\Phi(u))\rangle^2}}{|\alpha'(\Phi(u))|^3}=$$
$$\frac{\sqrt{|\alpha''(t)|^2|\alpha'(t)|^2-\langle \alpha'(t),\alpha''(t)\rangle^2}}{|\alpha'(t)|^3}$$
