The resistor from pin 7 is not limiting the first transistor emitter current, it's in parallel with the following b-e junction, so will actually increase the current a little. I think it's purpose is to bleed of any residual charge on the 2nd base when the devices turn off. Without this, as the base voltage falls to below 0.7v and the device turns off, the base is at a high impedance to the emitter, and any remaining charge has nowhere to go. The 1k ensures the base goes right to zero in good time.
I've encountered problems with transistor circuits that omit this, it can lead to unreliable operation and noise susceptability. Anything I design always has something to address this, it doesn't need to be as low as 1k, only a tiny current is required to quickly bleed off this charge, I usually use 100k or so if there isn't already some resistance from base to emitter.
I hope that helps.