Infinite differentiability and power series expansion Does every infinitely differentiable function have a power series expansion?Is this a theorem? Or is this an open question?

Nope!
Well, more precisely, for an infinitely differentiable function $f$, you can construct the Taylor series $t_a$ about the point $a$, but for most such functions, the equation
$$ f(x) = t_a(x) $$
only holds when $x=a$. (even when the radius of convergence of the Taylor series is nonzero!)
Functions that are actually equal to their power series around a point $a$ are called analytic at $a$. If they are analytic at all points, we call them analytic functions.  Most of the functions a mathematician works with do turn out to be analytic or at least piecewise analytic.
The traditional counterexample is
$$ f(x) = \begin{cases} 0 & x \leq 0
\\ e^{-1/x^2} & x > 0 \end{cases} $$
which is infinitely differentiable at $x=0$: in fact, $f^{(n)}(0) = 0$ for all natural numbers $n$. Thus its Taylor series is the zero function, which $f(x)$ is clearly not equal to.
This function is analytic everywhere else, though: i.e. if you computed a Taylor series about $1$ or $-1$, it would be equal to $f$ on an interval of positive radius. But this example makes for a good building block for constructing more complicated counter-examples.
