infinite differentiability of holomorphic functions
On the real line, having one derivative tells you almost nothing about having a second. The function x^2 sin(1/x) (suitably patched at 0) is differentiable once but not twice; you can build functions differentiable exactly k times and no more. In the complex world this whole ladder collapses to a single rung: a function that is complex-differentiable even once on an open set is automatically differentiable infinitely many times there.
This is a direct gift of the generalized Cauchy formula. Because f^(n)(z_0) = (n! / (2 pi i)) times the integral over gamma of f(z) / (z - z_0)^(n+1) dz exists for every n the moment f is holomorphic (Cauchy's formula only needed one derivative's worth of structure, namely holomorphy on and inside gamma), every higher derivative not only exists but is itself given by a contour integral — and a function given by such an integral is again holomorphic. So f', f'', f''', and so on are all holomorphic too. There is no 'twice but not thrice differentiable' holomorphic function.
Why care: this is the cleanest sign that complex differentiability is a far stronger condition than its real cousin. It is the bridge to analyticity — once you know f is smooth and its derivatives are controlled (Cauchy estimates), you can assemble its Taylor series and prove it converges to f. Holomorphic, infinitely differentiable, and analytic turn out to be three names for the same class of functions, an equivalence with no analogue for real functions.
The real function defined as x^2 sin(1/x) for x not 0 and 0 at x = 0 is differentiable once but its derivative is not differentiable at 0; no holomorphic function can behave this way — being complex-differentiable once forces all derivatives.
A real one-derivative-only example that has no complex analogue.
A common slip is to think this follows from the Cauchy-Riemann equations alone — it does not; you need holomorphy on a whole open set, which lets the integral formula act on a surrounding contour.