That's what I've got:
We know, that the region of absolute stability are those $z \in \mathbb{C}$, for which $|R(z)| < 1$, where $R(z)$ is the stability function.
For explicit Runge-Kutta methods, this is a polynomial, to be more precise, if we have an explicit RK method of p-th order, we get
$$R(z) = 1 + z + \frac{z^2}{2} + ... + \frac{z^p}{p!} + \mathcal{O}(z^{p+1})$$
So what I did is:
$$|R(z)| = |1 + z + \frac{z^2}{2} + ... + \frac{z^p}{p!} + \mathcal{O}(z^{p+1})| \le 1 + |z| + \frac{|z|^2}{2} + ... + \frac{|z|^p}{p!} + |\mathcal{O}(z^{p+1})|$$
But this means, that $|R(z)| < 1$ iff. $|z| + \frac{|z|^2}{2} + ... + \frac{|z|^p}{p!} + |\mathcal{O}(z^{p+1})| < 0$, which is simply impossible!
So what should I do now? Where's the mistake?
Edit: I thought, that maybe the other way would be to find out the region for which $z + \frac{z^2}{2} + ... + \frac{z^p}{p!} + \mathcal{O}(z^p) < 0$, but I don't think this could work as $z \in \mathbb{C}$, so unless I take the absolute value, this does not make sense for $\mathbb{C}$
So first of all, your mistake is that $\vert R(z) \vert$ is not less equal 1 if and only if
$$ \sum_{j=0}^p \left \vert \frac{z^j}{j!} \right \vert + \left \vert \mathcal O (z^{p+1}) \right \vert \leq 0.$$
Consider the simple counterexample of Forward Euler with stability polynomial $1+z$. Then, for $z=-1$ you have that $| R(z) | = 0$ while $| z | = 1 > 0$.
The reasoning behind the fact that the region of absolute stability is always bounded for explicit Runge-Kutta methods follows from the observation that for any non-constant polynomial $R(z)$ we have that $$ \lim_{| z | \to \infty} | R(z) | = \infty $$ Thus, the region of absolute stability is not unbounded = bounded.