Proof[edit]

Define the set on which ${\displaystyle f}$ and ${\displaystyle g}$ have the same Taylor expansion: $${\displaystyle S=\left\{z\in D\mid f^{(k)}(z)=g^{(k)}(z){\text{ for all }}k\geq 0\right\}=\bigcap _{k=0}^{\infty }\left\{z\in D\mid \left(f^{(k)}-g^{(k)}\right)(z)=0\right\}.}$$


We'll show ${\displaystyle S}$ is nonempty, open, and closed. Then by connectedness of ${\displaystyle D}$, ${\displaystyle S}$ must be all of ${\displaystyle D}$, which implies ${\displaystyle f=g}$ on ${\displaystyle S=D}$.


By the lemma, ${\displaystyle f=g}$ in a disk centered at ${\displaystyle c}$ in ${\displaystyle D}$, they have the same Taylor series at ${\displaystyle c}$, so ${\displaystyle c\in S}$, ${\displaystyle S}$ is nonempty.


As ${\displaystyle f}$ and ${\displaystyle g}$ are holomorphic on ${\displaystyle D}$, ${\displaystyle \forall w\in S}$, the Taylor series of ${\displaystyle f}$ and ${\displaystyle g}$ at ${\displaystyle w}$ have non-zero radius of convergence. Therefore, the open disk ${\displaystyle B_{r}(w)}$ also lies in ${\displaystyle S}$ for some ${\displaystyle r}$. So ${\displaystyle S}$ is open.


By holomorphy of ${\displaystyle f}$ and ${\displaystyle g}$, they have holomorphic derivatives, so all ${\displaystyle f^{(n)},g^{(n)}}$ are continuous. This means that ${\displaystyle \{z\in D\mid (f^{(k)}-g^{(k)})(z)=0\}}$ is closed for all ${\displaystyle k}$. ${\displaystyle S}$ is an intersection of closed sets, so it's closed.