BJT Voltage Divider Bias Circuit problem
Consider the equation that uses the reverse saturation current 

$$I_C=I_S\cdot e^\frac{V_{BE}}{V_T}$$

And then consider what happens for a slight difference in Vbe: -


If Vbe = 0.70 volts then Ic = 0.394 mA
If Vbe = 0.68 volts then Ic = 0.183 mA
If Vbe = 0.72 volts then Ic = 0.851 mA


So choose Vbe carefully or you'll be a mile out.

If you chose Vbe to be 0.72597 volts, Ic would equal 1.07 mA and match your 2nd derivation. It's easy to get fixated on what you believe are correct numbers.

As for the 2nd derivation using Kirchoffs, there is an omission that is relevant and will slightly lower the collector current. The omission I refer to is called \$r_E\$ and is the internal emitter resistance. 

In simple terms it equals \$\dfrac{V_T}{I_C}\$ 

Or about 26 ohms for an \$I_C\$ of about 1 mA and is fairly significant when you consider that \$R_E\$ is only 400 ohms. So, it's best not to get too fixated on some formulae. I'd be interested to know what a sim produces.