Suppose $G$ is a group, such that $\bigcap_{n=1}^\infty G^{(n)} = E$, where $G^{(n)}$ is the n-th element of the derived series of $G$, and $E$ is the trivial subgroup of $G$.
Does $G$ always have to be solvable?
For finite groups the answer obviously is "yes", but the situation with infinite groups seems to be much more complicated.
No, such a group need not be solvable. To get an example, for each $i$ let $G_i$ be a group of derived length $i$.
Then the restricted direct product $\prod_{i=1}^{\infty}G_i$ satisfies the condition that $\cap_{i=1}^{\infty}G^{(i)} = \{e\}$, but there is no $n$ such that $G^{(n)} = \{e\}$. Here by restricted direct product I mean where we only consider those sequences with finite support.