The base current in the BJT is due to the recombination of holes and electrons. To neutralize the p region an electron leaves the p region. If we take it the other way round, the larger the base current the greater is the recombination, so the collector current should reduce, but \$\beta I_c/I_b\$ . \$I_c=\beta I_b\$, which shows that \$I_b\$ is directly proportional to \$I_c\$.
First, if you study the literature, there are a lot of explanations of this, at every level. The one I liked best was the one I got in a class on solid-state quantum mechanics that was aimed at engineers, but the "earlier" ones (such as the one in the ARRL Handbook, just about any year after about 1970).
This phenomenon happens because the base is thin, and the geometry of the transistor is arranged so that the emitter is smaller than the collector. Ideally, the collector surrounds the emitter, with a thin shell of base between them.
When you put a voltage on the base (or put current into the base, which induces a voltage), this voltage causes current to flow through the base-emitter junction, pretty much as you'd expect for a diode. When there is sufficient voltage on the collector, instead of these carriers staying in the base and getting recombined, they go right through the base and into the collector.
The result of this is the BJT behavior that we know and love: you put a little bit of current into the base, and you get a lot of current going into the collector.