NPN BJT base->emitter resistance is effectively zero?
Understanding transistors is a bit like peeling an onion- there are many layers. At the simplest large-signal level you can consider the transistor as a current sink that's controlled by the current through the base-emitter junction. The latter behaves like a forward-biased diode. Not much current until you get to some hundreds of mV, and way too much current if you put volts across the junction. As you say, the transistor will conduct excessive current and will be destroyed if you simply connect (say) 5V to the base with emitter grounded. This is in stark contrast to the behavior of a MOSFET. 

At a more sophisticated level of understanding (which is required if you want to predict how most amplifiers work) and for small signals the base-emitter junction behaves like a resistor of Vt/Ib where Vt is the thermal voltage, about 26mV at room temperature. So if your base current is 2.5uA (say the beta is 300 and the transistor is biased with 0.75mA collector current), the base-emitter junction looks like about a 10K resistor for small signals. You can consider the transistor as a (somewhat imperfect because of r0) voltage controlled current source with an input resistance of Vt/Ib. This is the hybrid-\$\pi\$ model. Note that the transconductance gm (and thus the voltage gain in a common emitter configuration) is a function of the collector bias current and temperature and beta does not enter into it at all. 



I must emphasize that this model is a linearized model about a bias point and is quite invalid if the (change in) input voltage is large (more than some millivolts). In other words we're talking about relatively small changes on top of a fixed base-emitter voltage of perhaps 600 or 700mV.