is partial derivative holomorphic? Let $f : \mathbb{C} \to \mathbb{C}$ be a holomorphic function. If $z = x + \Bbb i y$, are the partial derivatives $\dfrac {\partial ^{k+l} f} {\partial x^k \partial y ^l}$ holomorphic functions themselves?

My idea: Proof by induction
$k=1: \frac{\partial f}{\partial x}$ is real differentiable. But why is $\frac{\partial f}{\partial x}$ complex differentiable?
 A: Yes, the partial derivatives of $f$ are themselves holomorphic, and this is easily shown using the Cauchy-Riemann relations. We want to prove that $\Bbb i \dfrac \partial {\partial x} \dfrac {\partial ^{k+l} f} {\partial x^k \partial y ^l} = \dfrac \partial {\partial y} \dfrac {\partial ^{k+l} f} {\partial x^k \partial y ^l}$, knowing that $\Bbb i \dfrac {\partial f} {\partial x} = \dfrac {\partial f} {\partial y}$. Using the symmetry of the higher-order partial derivatives we may write
$$\Bbb i \dfrac \partial {\partial x} \dfrac {\partial ^{k+l} f} {\partial x^k \partial y ^l} = \Bbb i \dfrac {\partial ^{k+1+l} f} {\partial x^{k+1} \partial y ^l} = \dfrac {\partial ^{k+l}} {\partial x^k \partial y ^l} \Bbb i \dfrac {\partial f} {\partial x} = \dfrac {\partial ^{k+l}} {\partial x^k \partial y ^l} \dfrac {\partial f} {\partial y} = \dfrac {\partial ^{k+l+1} f} {\partial x^k \partial y ^{l+1}} = \dfrac \partial {\partial y} \dfrac {\partial ^{k+l} f} {\partial x^k \partial y ^l}$$
which is exactly what we want to get.
