Two different ways to define a Kähler metric on a complex manifold are:
1) The fundamental form $\omega = g(J\cdot,\cdot)$ is closed, ie, $d\omega=0$;
2) The complex structure $J$ is parallel with respect to the Levi-Civita connection of $g$, ie $\nabla J=0$.
Using the formula
$$d\omega(X,Y,Z) = X\omega(Y,Z) - Y\omega(X,Z) + Z\omega(X,Y) - \omega([X,Y],Z) + \omega([X,Z],Y) - \omega([Y,Z],X),$$
I could easily prove that $\nabla J = 0$, i.e. $\nabla_XJY=J\nabla_YX$.
What I'm trying to understand is why $d \omega=0 \Rightarrow \nabla J =0$. There's a proof on Kobayashi-Nomizu vol. 2, a consequence of the formula
$$4g((\nabla_X J)Y,Z) = 6 d \omega (X,JY,JZ) - 6 d \omega(X,Y,Z) + g(N(Y,Z),JX)$$
but it mentions the Nijenhuis tensor $N$. I'm looking for a proof that doesn't use that. I think a proof is possible by using the Koszul formula for the Levi-Civita connection and the formula for $d\omega$ above wisely used. Does anyone knows a proof that doesn't mention $N$ explicitly?