BJT small signal analysis, ignoring current sources
The current supplied 2nd source \$g_{m2}v_{\pi 2}\$ is not negligible and is not ignored here. 

The current entering to node \$v_{b2}\$ from left is the base current to second transistor (let it be \$i_{b2}\$). Then the collector current will be \$\beta_2i_{b2}\$ hence the emitter current will be \$(\beta_2 + 1)i_{b2}\$. That is why the final equation has a multiplication term of \$(\beta_2 + 1)\$ in it.

OR the 2nd source provides a current of \$\beta_2i_{b2}\$.