Harmonic conjugate on a simply connected domain existence proof with the skew-gradient

Question

I am a physics student following some notes on complex analysis, specifically their relation to harmonic functions. There is a proof that every harmonic function defined on a simply connected domain has a harmonic conjugate that forms, with it, an analytic complex function on the same domain (or so I understand). The proof involves using the skew-gradient $$\nabla v=\nabla^{\perp}u\tag{1},$$ with $$\nabla^{\perp}=\begin{pmatrix}-\partial_y \\ \partial_x\end{pmatrix}\tag{2}.$$ We then use the fact that a potential function exists for $\nabla^{\perp}u$ iff a line integral in the domain is path independent. The following step does something that confuses me: $$0=\oint_\mathcal{C}\vec\nabla v\cdot d\vec x=\oint_\mathcal{C}\vec\nabla^{\perp}u\cdot d\vec x=\oint_\mathcal{C}\vec \nabla u\cdot \vec n\text{ }ds.$$ The last step is confusing to me, we are saying that any closed integral must be $0$. We then take the closed integral of the gradient field of the function $v$ around $\mathcal{C}$. This is equal to the skew-gradient field of $u$, that makes sense since they're the same thing by definition. But in the last step we introduce a new vector $\vec n$, which I assume is some normal vector, but to which direction? Any help is appreciated, thank you.