Holomorphic functions are infinitely differentiable

Statement

Let $U \subseteq \mathbb{C}$ be an open set and let $f : U \to \mathbb{C}$ be holomorphic on $U$. Then the complex derivative $f'$ is holomorphic on $U$. Thus, for each $n \ge 1$, the $n$-th complex derivative $f^{(n)}$ exists and is holomorphic on $U$. As a function of two real variables $(x, y)$ with $z = x + iy$, $f$ has continuous partial derivatives of all orders on $U$.

Remarks

Proof: near each point, $f$ is the sum of a power series by holomorphic functions are analytic. A power series can be differentiated term by term, and the result is again a power series. There is no real analogue: $x \mapsto x |x|$ is differentiable on $\mathbb{R}$, but its derivative $2|x|$ is not differentiable at $0$. The formula for the derivatives gives each $f^{(n)}$ as a circle integral.

Explanations

L1 human

Once a complex function has one derivative on an open set, it has a second one, a third one, and so on forever. So in complex analysis there are no functions that are differentiable only once.