I am unable to see why the following statement is trivial:
Since $\mathbb{C}^n$ is contractible, we see that $\check{H}^k(\mathbb{C}^n,\mathbb{Z}) = 0$ for $k>0$.
Source: p. 46, "Principles of algebraic geometry" - Griffiths and Harris.
A similar question was asked here earlier also, but it doesn't contain a satisfactory answer.