I have trouble with understanding Shabat's proof of Hartogs' Lemma, used in a proof of Hartog's theorem. Let $U(a,r)$ denote polydisc centered at $a$ with radii $r$, $'z=(z_1,...,z_{n-1})\in\mathbb{C}^{n-1}$, $'0=(0,...,0)\in\mathbb{C}^{n-1}$, $'V=U('0,R), 'W=('0,r), 0<r<R, U_n=\{|z_n|<R\},V='V\times U_n,W='W\times U_n$. Let $f('z,z_n)$ be holomorphic in $\overline{'V}$ with respect to $'z$ for every $z_n\in\overline{U_n}$ and holomorphic with respect to $z=('z,z_n)\in\mathbb{C}^n$ in $\overline W$.
For every $z_n\in U_n$ and $'z\in\,'V$ $f$ can be represented as a convergent multiple power series $$f(z)=\sum_{|k|=0}^\infty c_k(z_n)('z)^k$$ where $k=(k_1,...,k_n), |k|=k_1+...+k_n,\,'z^k=z_1^{k_1}\cdots z_{n-1}^{k_{n-1}}$. Shabat claims that the coefficients $$c_k(z_n)=\frac1{k!}\frac{\partial^{|k|}f('0,z_n)}{(\partial\,'z)^k}$$ are holomorphic in $U_n$ as derivatives of a holomorphic function wrt $z_n$. I fail to see why. We differentiate wrt $'z$, $f('z,z_n)$ is holomorphic in $'V$ with $z_n$ fixed, I don't see how to obtain holomorphy of $c_k(z_n)$ in $U_n$.