Proof of (0x66e293b5)

We prove (0x66e293b5) by an induction argument. For \(n=1\) see (0x66e28ddc) . Assume the statement is true for some \(n\in \mathbb{N}\). Then

\begin{align*} (f\circ \gamma)^{(n+1)}&=((f\circ \gamma)^{(n)})'\\ &=(\nabla^n_{\gamma'}f)'\\ &=\nabla_{\gamma'}(\nabla^n_{\gamma'} f)\\ &=\nabla^{n+1}_{\gamma'} f, \end{align*}

where we sequentially used the induction assumption, the formula in (0x66e28ddc) and (0x66e29b2a) .