Why are differentiable complex functions infinitely differentiable?
As Akhil mentions, the keyword is elliptic regularity. Since I don't know anything about this, let me just say some low-level things and maybe they'll make sense to you.
A differentiable function $f : \mathbb{R} \to \mathbb{R}$ can be thought of as a function which behaves locally like a linear function $f(x) = ax + b$. So, very roughly, it is a collection of tiny vectors which fit together. These tiny vectors can, however, fit together in a very erratic manner. That's because since you only have to fit one vector to the two vectors that are its neighbors, there is a lot of room for bad behavior.
A differentiable function $f : \mathbb{C} \to \mathbb{C}$ has to satisfy a much more stringent requirement: locally, it has to behave like a linear function $f(z) = az + b$ where $z, a, b$ are complex, which is a rotation (and scale, and translation). So, very roughly, it is a collection of tiny rotations which fit together. Now one rotation has a continuum of neighbors to worry about, and it becomes much harder for erratic behavior to persist.
The existence of a complex derivative means that locally a function can only rotate and expand. That is, in the limit, disks are mapped to disks. This rigidity is what makes a complex differentiable function infinitely differentiable, and even more, analytic.
For a complex derivative to exist, it has to exist and have the same value for all ways the "h" term can go to zero in $\frac {(f(z+h) - f(z))}{h}$. In particular, h could approach 0 along any radial path, and that's why an analytic function must map disks to disks in the limits.
By contrast, an infinitely differentiable function of two real variables could map a disk to an ellipse, stretching more in one direction than another. An analytic function can't do that.
A smooth function of two variables could also flip a disk over, such as $f(x, y) = (x, -y)$. An analytic function can't do that either. That's why complex conjugation is not an analytic function.
When one uses the complex plane to represent the set of complex numbers ${\bf C}$,
$z=x+iy$
looks so similar to the point $(x,y)$ in ${\bf R}^2$.
However, there is a difference between them which is not that obvious. The linear transformation in ${\bf R}^2$, can be represented by a $2\times 2$ matrix as long as one chooses a basis in ${\bf R}^2$, and conversely, any $2\times 2$ matrix can define a linear transformation by using the matrix multiplication $A(x,y)^{T}$.
On the other hand, the linear transformation on $\bf C$ is different. Let $f:{\bf C}\to{\bf C}$ where $f(z)=pz$, $p \in{\bf C}$. If one writes $p=a+ib$ and $z=x+iy$, this transformation can be written as
$$ \begin{bmatrix} x\\ y \end{bmatrix}\to \begin{bmatrix} a &-b\\ b &a \end{bmatrix} \begin{bmatrix} x\\ y \end{bmatrix} $$
when one sees it as in the complex plane. Hence, not all matrices can define a linear transformation $f:\bf C\to C$.
The derivative, which can be regarded as a "linear transformation", is also different for $f:{\bf R}^2\to {\bf R}^2$ and $f:\bf C\to C$. In the real case
$$ f \left( \begin{bmatrix} x\\ y \end{bmatrix} \right) = \begin{bmatrix} f_1(x,y)\\ f_2(x,y) \end{bmatrix} $$
$f_1$ and $f_2$ are "independent" for the sake of $f$ being differentiable.
While in the complex case $f_1$ and $f_2$ have to satisfy the Cauchy-Riemann equations.
The relationship between $f:{\bf R}^2\to{\bf R}^2$ and $f:{\bf C}\to{\bf C}$ is also discussed here.