Let $E\rightarrow M$ be a vector bundle with metric $g$ and metric connection $\nabla:\Omega^0(M,E)\rightarrow \Omega^1(M,E)$.
I am trying to understood this short proof:
I do not understand the second equality. How does this come from the Leibniz rule?
I think the second equality isn't so much the Leibniz Rule; really that equality is the definition of the connection $\nabla$ induces on $\wedge^2 E^*$. I will explain. If $s, t$ are sections of $E$, then $$ \begin{align*} (\nabla \omega)(s,t) &:= d(\omega(s,t)) - \omega(\nabla(s),t) - \omega(s,\nabla(t))\\ &= d(g(F_\nabla s, t)) - g(F_\nabla \nabla s,t) - g(F_\nabla s, \nabla t)\\ &= (\nabla g)(F_\nabla s,t) + g(\nabla (F_\nabla s), t) + g(F_\nabla s, \nabla t) - g(F_\nabla \nabla s,t) - g(F_\nabla s, \nabla t)\\ &= (\nabla g)(F_\nabla s,t) + g(\nabla (F_\nabla s), t) - g(F_\nabla \nabla s,t) \\ &= (\nabla g)(F_\nabla s,t) + g(\nabla (F_\nabla s)-F_\nabla \nabla s, t)\\ &= (\nabla g)(F_\nabla s,t) + g((\nabla F_\nabla)s, t). \end{align*} $$ In the first line I just used the definition of the induced connection. The second line is inserting the definition of $\omega$. The third line I rewrote $d(g(-,-))$ as $(\nabla g)(-,-) + g(\nabla-,-) + g(-,\nabla -)$, which is another use of the definition of the induced connection. The fourth line I canceled the terms both involving $g(F_\nabla s, \nabla t)$. Fifth line is linearity of $g$, and the sixth line is the definition of $\nabla(F_\nabla)$. This gives the formula you're after.
It seems to me they use the Liebniz rule to show that $\nabla(\omega^k) = 0$ once the case $k=1$ is established.