It's a bit unclear what you are really asking, but it seems the answer is "Because that's what transistors do". For more details, see semiconductor physics.
Also, you should be thinking about the current thru the base, not so much the B-E voltage as driving things. B to E looks like a diode to the external circuit, so will have about 700 mV drop depending on current. But, the function of voltage to current is quite steep, and current is ultimately what matters anyway.
V2 in your schematic should be a current source, not a voltage source. Let's say you set it for 100 µA and the transistor has a gain of 50. That means the collector can sink up to (100 µA)50 = 5 mA. If R1 were 300 Ω, then it would drop (5 mA)(300 Ω) = 1.5 V. The collector would therefore be at 3.5 V.
The above example was assuming the transistor had a gain of exactly 50 at the operating point we chose. Real transistor gains vary widely. The datasheet generally specifies a minimum, but no maximum. It's not unheard of for actual parts to have 10x more gain than the minimum guaranteed value.
Unpredictable gain is one reason transistors used in linear operation are biased with some feedback. The actual C-E voltage is used to adjust the bias current to keep the collector voltage within a useful range over the possible range of gains from part to part.