How to calculate Ic for a very simple NPN transistor circuit
In your circuit, you are still not driving the BJT into deep saturation region (switching behavior), since the base voltage (760mV) is slightly larger than the diode's forward voltage (650mV). 

If you sweep the base voltage, you would get the following curve. Here you still see that the output current still has some strong dependency on the base voltage range between 700mV and 900mV (your case). I might be wrong here, but the exponential growth of the current comes from the fact, that you do not have any base resistor and the current is mostly a function of the emitter's diode current (formula that you wrote in your question).



If you do not add any collector resistor, the collector current will be given by:

$$I_C = I_B \cdot \beta = I_S \cdot (e^{\frac{V_D}{n \cdot V_T}}-1) \cdot \beta$$

More to that see this link.