I think that one possible answer is that a Calabi-Yau manifold is a Riemannian manifold $M$ with $SU(n)$ Riemannian holonomy, where $n=2\dim_\mathbb R M$.

Such a manifold is then necessarily complex, and the Riemannian metric is the real part of a Kähler metric which has zero Ricci curvature. Since the Ricci form in complex geometry is always a representative of the first Chern class of the manifold, what you ask follows.

I think that one possible answer is that a Calabi-Yau manifold is a Riemannian manifold $M$ with $SU(n)$ Riemannian holonomy, where $2n=\dim_\mathbb R M$.

Such a manifold is then necessarily complex, and the Riemannian metric is the real part of a Kähler metric which has zero Ricci curvature. Since the Ricci form in complex geometry is always a representative of the first Chern class of the manifold, what you ask follows.