A couple of the other answers have rearranged the circuit diagram into a more reasonable form for you, which is helpful and makes it much easier to reason about -- you can see that it is a simple circuit of series and parallel resistors, which you probably know rules to simplify.
But I actually find this aspect your question interesting: how do we know that the current does not get trapped in a loop? It is obvious to me, and to the other answerers here, that it won't; but what's the easiest way to show that for certain?
In a very simple circuit like this, made of only resistors, you can reason as follows: for some fixed voltage across points A and B (OR equivalently some fixed current flowing in at A and out at B), the circuit will be in an equilibrium with currents flowing through the various resistors, and voltages being present at the various points where they connect to each other. (More properly, there are voltage differences that you could measure between pairs of such points, but in most cases, including this circuit, you can get away with acting like there are absolute voltage values at the nodes.)
Now, without seeing any numbers at all, we can say this: For any resistor in this circuit, if there is a voltage difference across it that is not zero, current will flow; and it will flow from the higher voltage side to the lower.
Now, can you assign voltages to the nodes that will make current flow in a loop? Again, this is obvious to anybody who works with circuits a lot, but it's worth thinking about why you can't. One way to see it: consider the loop formed by R2-R3-R4-R8. There are four nodes forming that loop (at the points where the resistors connect), and each has a voltage. Assuming none are the same, one must be highest, and one must be lowest, out of the four. Then current cannot flow in a loop, because it must flow outwards both ways from the highest node, and inward both ways to the lowest node.