I saw the proof of $$\mathrm{one-form} \ \omega \ \mathrm{is \ exact} \Rightarrow \int_{S^1} \omega =0$$ where $S^1$ is an unit circle.
The proof is here.
Since $\omega$ is exact, there exists $f\in C^\infty$ s.t. $df=\omega.$
Then, $\int_{S^1} \omega=\int_{S^1}df=\int_{\partial S^1} f=0.$
I know that $\int_{S^1}df=\int_{\partial S^1}f$ is lead from Stokes, but I don't know why $\int_{\partial S^1} f=0$ holds.
I want you to explain why this equality holds.
Roughly speaking, an interval $[a,b]$ with $a<b$ has two boundary points $a$ and $b$. $S^1$ is like taking an interval and making it close back on itself, so the endpoints $a,b$ "become the same", so $S^1$ has no more boundary points. More formally, $S^1$ is a smooth manifold in the usual sense (every point of $S^1$ has an open neighborhood and a chart map, which maps it bijectively onto an open subset of $\Bbb{R}$). The fact that $S^1$ is a smooth manifold is surely something you must have seen much earlier than Stokes theorem.
Note that in a manifold $M$, possibly having boundary, if a point $p\in M$ is such that it has an open neighborhood along with a chart mapping to an open subset of $\Bbb{R}^n$ (rather than the half space $\Bbb{H}^n$), then $p$ is NOT part of $\partial M$ (this is saying that the manifold boundary $\partial M$ the manifold "interior" are disjoint sets; this can be proven relatively easily using the inverse function theorem).
Hence, in the case of $S^1$, it has no boundary points (since we showed directly that every point is an "interior" point). Hence, $\partial S^1=\emptyset$.