Proof of the second Bianchi identity I'm asked to prove the second Bianchi identity:
$$\nabla_{[e}R_{ab]c}^{\;\;\;\;d}=0$$
using the fact that:
$$(\nabla_a \nabla_b -\nabla_b \nabla_a)\omega_c=R_{abc}^{\;\;\;d}\omega_d$$
For every diff. form $\omega$.
I can't.
Using that I'm able to reduce the lhs of the identity to:
$$2\nabla_{[e}\nabla_a \nabla_{b]}\omega_c$$
But I can't go further.
 A: It is the ambiguity of the conventional abstract index notation for the covariant derivative that causes troubles here. Getting used to it over time helps :-)
We say that a tensor $T_{abc\dots}$ has the Bianchi symmetry if $T_{[abc]\dots} = 0$.
Your goal is to show that the tensor $\nabla_e R_{abc}{}^d := (\nabla R)_{eabc}{}^d$ has the  Bianchi symmetry in the indices $e,a,b$, and this fact is called the Bianchi identity. 
The fact 
$$
2 \nabla_{[a} \nabla_{b]} \omega_c = (\nabla_a \nabla_b -\nabla_b \nabla_a)\omega_c=R_{abc}^{\;\;\;d}\omega_d
$$ 
can be seen as the definition of the curvature $R$.
Taking the covariant derivative of $R_{abc}{}^d \omega_d$ by an application of the Leibniz rule we see that
$$
\nabla_e (R_{abc}{}^d \omega_d) =  (\nabla_e R_{abc}{}^d) \omega_d +  R_{abc}{}^d \nabla_e \omega_d
$$
which means that for any $\omega_d$ the tensor $\nabla R$ satisfies the identity
$$
(\nabla_e R_{abc}{}^d) \omega_d  = \nabla_e (R_{abc}{}^d \omega_d) - R_{abc}{}^d \nabla_e \omega_d
$$
Now we can use the definition of $R$ and rewrite the last display as
$$
(\nabla_e R_{abc}{}^d) \omega_d  = 2 \nabla_e (\nabla_{[a} \nabla_{b]} \omega_c) - 2 \nabla_{[a} \nabla_{b]} \nabla_e \omega_c + R_{a b e}{}^d \nabla_d \omega_c
$$
where we have used the Ricci identity
$$
2 \nabla_{[a} \nabla_{b]} t_{ec} = R_{a b e}{}^d t_{d c} +  R_{a b c}{}^d t_{e d}
$$
I will wait now for you to figure out that we are almost done.
