A map $f:U\ni(z_1,\dots,z_n)\mapsto f(z_1,\dots,z_n)\in\mathbb C$ (with $U\subseteq\mathbb C^n$ open) is called holomorphic if $f$ is real continuously differentiable and for every $i\in\{1,\dots,n\}$, we have $\frac{\partial f}{\partial\bar{z_i}}=0$.
Hartogs' Theorem now states that $f$ is holomorphic if and only if $f$ is holomorphic with respect to every variable $z_1,\dots,z_n$, see here.
I don't see why this is actually a Theorem. Is this not just a reformulation of the definition? Or is it that the definition that I take for holomorphicity is not the one that Hartogs used?