The fundamental vector fields of a principal bundle are vertical. Let $p:P\to M$ be a principal $G$-bundle. To each $A$ in the Lie algebra of $G$ corresponds a fundamental vector field $A^*$ on $M$ defined by
$$A^*_u=\frac{d}{dt}|_{t=0} u(exp(tA))$$
How can we see that $A_u^*$ is a vertical vector?
Idea: We need to verify that $p_*(A_u^*)=0$. None of the manipulations that I apply to the left hand side help.
Supposedly, the statement is obvious from the fact that the action of $G$ takes each fiber to itself.

First of all a fundamental vector field is a vector field on $P$ not on $M$ as you wrote in your question. As you say the action of $G$ on $P$ preserves the fibers i.e. $p(u) = p(u g)$ for all $u \in P , g \in G$. If $A$ is in the Lie algebra of $G$ then $exp(tA)$ is a monoparametric group. For $u \in P$ we have a curve $\gamma(t):= u.exp(tA)$ in $P$. By the assumption the curve $\gamma$ is contained in a fiber: $$p(\gamma(t)) = p(\gamma(0))$$ 
and taking derivative you get $$p_*(A_u^*) = 0$$
which show you that indeed $A_u^*$ is vertical i.e. tangent to a fiber.
