\$i_{b1}\$ in this simple bjt circuit
The output voltage of this circuit is equal to minus the collector current flowing in the collector resistance times the collector resistance: \$V_{out}=-i_cR_C\$ as shown in the below sketch:


The base current is equal to the voltage across \$r_\pi\$ divided by \$r_\pi\$. That is \$i_b=\frac{V_{in}-V_E}{r_\pi}\$. The emitter voltage is the drop across the emitter resistance \$R_E\$ in which a current made of the sum of the base current \$i_b\$ with the collector current \$i_c\$ flows. Therefore, \$V_E=R_E(i_b+i_c)=R_E(i_b+\beta i_b)=R_E(\beta+1)i_b\$. Now extract \$i_b\$ from this definition and substitute it in the first expression we derived. You should have \$i_b=\frac{V_{in}}{(\beta+1)R_E+r_\pi}\$. You know that the output voltage \$V_{out}\$ is equal to \$-i_cR_C=-\beta i_bR_C\$. Substitute the last definition of \$i_b\$, rearrange and you should obtain \$\frac{V_{out}}{V_{in}}=-\frac{R_c\beta}{r_\pi+R_E(\beta+1)}\$. If the transistor gain \$\beta\$ is large enough, this expression simplifies to \$\frac{V_{out}}{V_{in}}\approx -\frac{R_C}{R_E}\$.

A quick dc bias point simulation with a Mathcad sheet shows the result is correct: