According to Chern-Weil theory Chern forms $c_i$ of the vector bundle $\xi : E \to M$ are determined by the polynomial $$ \det\left(I + \frac{\mathrm{i}t}{2\pi}F \right) = 1 + \sum^n_{i=1} c_i(\xi) t^i, $$ where $F$ is a curvature form determined by the connection form $\omega$ as $$ F = \mathrm{d}\omega + [\omega \wedge \omega]. $$ Now Chern class is the homological class $[c_i(\xi)]$.
It is evident that for the line bundle $\xi$ and for any $i > 1$ it holds that $c_i(\xi)$ what I can't understand is that why $[c_1(\xi)] \neq 0$ for nontrivial line bundles.
In case of line bundle expression above simplifies to
$$ \det\left(I + \frac{it}{2\pi}F \right) = 1 + \frac{it}{2\pi} F $$ But on the line bundle $\omega$ is just a differential form, so $ F = \mathrm{d}\omega $ as $\omega \wedge \omega = 0 $. Hence, $c_1(\xi) = \frac{i}{2\pi} d \omega$ is an exact form, and so $[c_1(\xi)] = 0$. But this contradicts many results concerning line bundles!
It seems that I don't understand something either about definition of Chern class or about the cutvature form. Please help me find my mistakes.
The expression you wrote down in your argument for vanishing first Chern-class, would work if the bundle was trivial in the first place. This is exactly what Peter wrote.
Remember (assuming standard literature) that said expression is a result on local trivialisations, for $\omega$ the local connection form.
So your argument works locally, but that is of course not the point.
The real question is now: "why is vanishing first Chern-class equivalent to being trivial for line bundles?"