Let $A \in \mathbb R^{n \times n}$ and assume that there is an $\lambda \in \mathbb R$ such that $\langle x, A x \rangle \leq \lambda\langle x,x\rangle$ for all $x\in\mathbb R^n$. Show that $$\|\exp(t A)x\| \leq e^{\lambda t}\lVert x\rVert,\qquad t \geq 0$$
My problem here is that I don't know how to apply the given inequality when starting with $\lVert \exp(tA)x\rVert$. What can I do with that expression so that $A$ is not "bounded" in the exponential and I can use the inequality? Is it a certain property of the norm?
Edit: I have tried to plugin the exponential series:
$$||\exp(tA)x||\leq\sum_{n=0}^\infty\frac{t^n}{n!}||A^nx||\leq\dots\leq\sum_{n=0}^\infty\frac{t^n}{n!}\lambda^n||x||$$ but I haven't been able to fill out the $\dots$ so far.