I am trying to derive the formula for torsion of a curve. In many example proofs I've seen the final step is that $\kappa=\|\dot{\lambda}\times\ddot{\lambda}\|$. However I thought curvature was defined to be $\kappa=\frac{\|\dot{\lambda}\times\ddot{\lambda}\|}{\|\dot{\lambda}\|^3}$ or $\kappa=\|\ddot{\lambda}\|$ if $\lambda$ is unit-speed. However, I would like to note, I have only assumed that $\lambda$ is a regular curve with non-zero curvature.
Edit: essentially how does one derive the formula in the screenshot.
We must use the Frenet formulas for regular curves: $$ \begin{align*} T' &= v\kappa N\\ N' &= -v\kappa T + v \tau B\\ B' &= -v\tau N. \end{align*} $$ Here $T, N, B$ is the Frenet frame along the curve $\gamma$, $v = \|\gamma'\|$ is the speed and $\kappa$ and $\tau$ are of course the curvature and torsion.
By definition of $v$ we have $$ \gamma' = vT. $$ The second derivative of $\gamma$ is $$ \gamma'' = v'T + vT' = v' T + v^2\kappa N. $$ The cross product $\gamma'\times \gamma''$ becomes $$ \gamma' \times \gamma'' = \kappa v^3 B. $$ If we take the norm of both sides, we obtain $$ \kappa = \frac{\|\gamma'\times\gamma''\|}{v^3} = \frac{\|\gamma'\times \gamma''\|}{\|\gamma'\|^3}. \tag{1} $$ The binormal vector $B$ is a unit vector perpendicular to $T$ and $N$. From the expressions for $\gamma'$ and $\gamma''$, it follows that $B$ points in the same direction as $\gamma'\times \gamma''$, so $$ B = \frac{\gamma'\times\gamma''}{\|\gamma'\times\gamma''\|}. \tag{2} $$
Deriving one more time gives $\gamma'''$: $$ \begin{align*} \gamma''' &= v'' T + v'T' + (v^2\kappa)'N + v^2\kappa N' \\ &= v''' T + (v'v\kappa + (v^2\kappa)')N + v^3\kappa (-\kappa T + \tau B). \end{align*} $$ Take the inner product with $B$ and obtain $\gamma'''\cdot B = v^3\kappa \tau$. Combined with $(2)$ this gives $$ \tau = \frac{\gamma''' \cdot \gamma'\times\gamma''}{\|\gamma'\times\gamma''\|v^3\kappa}. $$ Finally, using $(1)$, we may conclude that the torsion is given by $$ \tau = \frac{\gamma'\times\gamma''\cdot\gamma'''}{\|\gamma'\times\gamma''\|^2}. $$