How can a transistor amplify current in a circuit?
I can understand your doubts - because, in reality, the transistor does NOT amplify the base current. It is true that the collector curent \$I_c\$ is proportional to the base current \$I_b\$ (\$I_c/I_b=\beta\$), but this is a kind of correlation.
(It is really a pity that there are still some books and publications claiming that \$I_b\$ would determine \$I_c\$. Nevertheless, during design and/or analysis of transistor stages we can in many cases treat the transistor as if \$I_b\$ would determine \$I_c\$; this is because the relation \$I_c=\beta I_b\$ does apply - but it is a correlation and does not reflect a causality).
It is not a problem to show and to verify that \$I_b\$ as well as \$I_c\$ are both dependent on the voltage \$V_{be}\$ according to Shockley's equation \$I_e = I_s[\mathrm{exp}(V_{be} / V_t)-1]\$ because the emitter current \$I_e\$ is split into a very small current (\$I_b\$) and a larger current \$I_c\$ (\$I_e = I_b + I_c\$).

Final (summarizing) statement (with respect to the long list of comments):
The following sentence alone cannot explain the transistor principle, but it shows that it is the VOLTAGE which plays the decisive role:
In every conductor/semiconductor, a current (movement of electric charges) can exist under the influence of an electric field only. In electronic circuits, this E-field is generated by an external voltage. A change of the voltage (of the E-field) causes a change of the current. This is true, of course, also for diodes, bipolar transistors (BJTs) and FETs.