exterior derivative vs connection: $d= \theta^i\wedge\nabla_{e_i}$

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Let $\nabla$ be the Levi-Civita connection of a Rimannian manifold $M$, let $e_1,\dots,e_n$ denote an orthogonal local flame field of $TM$ and let $\theta^1,\dots,\theta^n$ be the its dual coflame. I would like to show that $$d\alpha=\sum_{i=1}^{n} \theta^i\wedge\nabla_{e_i}\alpha $$ for any 1-form $\alpha$ (see wiki). To do this, let $\alpha=\sum_{i=1}^n\alpha_i\theta^i$ and $\theta^i=\sum_{j=1}^ng_j^idx^j$. Since the Levi-Civita connectin is torsion free, it follows $$d\theta^i=\sum_{j=1}^n\theta^j\wedge\omega_j^i$$ where $\omega_i^j$ is the connection form of $\nabla$. So $$\sum_i \theta^i\wedge\nabla_{e_i}\alpha=\sum_{i} \theta^i\wedge \left(d\alpha_i-\sum_j\alpha_j\omega_i^j\right) \\ =\sum_i\theta^i\wedge d\alpha_i-\sum_i\alpha_id\theta^i \\ =-\sum_i d\alpha_i\wedge\theta^i-\sum_i\alpha_id\theta^i \\ =-d\left(\sum_i\alpha_i\theta^i\right)=-d\alpha$$

Why is it incorrect?