Flat Chern curvature implies a trivial holomorphic vector bundle

Question

Let $(M,g)$ be a compact complex manifold, and let $\mathcal{V}$ be an Hermitian holomorphic vector bundle over $M$, with $\nabla$ its Chern connection. If the curvature operator $\nabla^2 = 0$, then is it necessarily true that $\mathcal{V}$ is a trivial vector bundle. I guess this might be proved using Hirzebruch--Riemann--Roch, and the knowledge that a non-trivial vector bundle should have non-trivial Euler characteristic. However, it seems there should be a much simpler way to do this . . .