Why does Vbe decreases with the increase in temperature for an n-p-n transistor?
It decreases because Is is an exponential function of temperature, roughly doubling to quintupling for every 10°C, depending on \$\eta\$. 

Much more from this chapter. 

The equation for change is derived as follows: 

\$dv/dT =  \frac{v - (3V_T +V_G)}{T}\$ 

where 
Vt is the thermal voltage 0.0259V @ 300K (Vt = kT/q)

Vg is the bandgap voltage for silicon 1.11V

T is the absolute temperature say 300K 

so at room temperature of 300K with v of 600mV 
dv/dT is about -1.96mV/°C