Let $\Omega$ be a simply connected domain, i.e. open and connected subset of $\mathbb{C}$ and $f$ be analytic in $\Omega$. I am trying to understand how you can deduce that $f(z) = \frac{d}{dz}F(z)$ for some analytic function $F:\Omega\to\mathbb{C}$ from the two following results (which are verbatim from my old notes handed in class):
1.) Monodromy theorem: If $\Omega$ is simply connected and $f:\Omega'\to\mathbb{C}$ is an analytic function which admits an analytic continuation along any path $\gamma$ in $\Omega$ with the starting point of $\gamma$ staying in $\Omega'$, then $f$ has a direct analytic continuation to $\Omega$.
2.) If $f$ is an analytic function in a domain $\Omega$ and $F$ is its primitive, i.e. $\frac{d}{dz}F(z) = f(z)$, then $F$ has an analytic continuation along $F_k:\Omega_k\to\mathbb{C}, k=1,\dots,n$, i.e. $F_{k+1}\equiv F_k$ in $\Omega_{k+1}\cap \Omega_k$, where $\Omega_k$s are discs and $\frac{d}{dz}F_k(z) = f(z)$ in $\Omega_k$.
(Thoughts:) My current understanding is nil in the sense that I do not see what Monodromy theorem has to do with the second result, which likely used to conclude the result. Specifically, the analytic continuation along $F_k$ of the second result almost fit with out need: Their derivative agrees with $f$ on the disks $\Omega_k$. To satisfy our case we would need to inflate some of (or all) the disks $\Omega_k$ so that inflated $\tilde{\Omega_k} = \Omega$. But I just don't see how this happens with the Monodromy theorem.
Edit: As discussed in the comments, the function $F(w) = \int_{\gamma_{w_0, w}}f(z)dz$ will do as such an analytic function where $w_0\in\Omega$ is fixed and $\gamma_{w_0, w}$ is a path contained in $\Omega$ connecting $w_0$ and $w$. But what is still unclear to me is that where do we need the two stated theorems? To me the only thing we need is to be able to conclude that in simply connected domains, the path integral of an analytic function depends only on the initial and terminal point.