Classic transistor differential ampliﬁer - problem with a single-ended output
OK for Vcc and Ic

As for the gain, each transistor behaves like it has an internal emitter resistor of value:

\$ r_e = \frac{1}{g_m} = \frac{V_T}{I_C} \$ with Vt about 25mV.

So, with zero input voltage, Ic=1mA per transistor (2mA total), and their re are both 25 ohms.

To calculate the gain we need \$ \frac{dIc}{dVin} \$ so we approximate both transistors as perfect transistors with emitter resistor re added. Since each transistor has its own re, they appear in series, so 

\$ \frac{dIc}{dVin} = \frac{1}{2 r_e} = \frac{I_C}{2V_T} \$

Multiply by Rc to get the gain...

\$ \frac{I_C Rc}{2V_T} \$

I think you lost a "2" somewhere along the way, as Ic=Io/2 my result doesn't match yours...

Note that the gain depends on transistor gm which depends on collector current for each transistor. If the collector currents are not balanced then the transconductance \$ g_m \$ of one transistor will increase, and the other will decrease. Same for their \$ r_e \$ since \$ r_e = 1/g_m\$

At low input voltages these variations compensate each other, and linearity is good. However when input voltage is high enough to produce a noticeable variation of current in the transistors, or even further, when one transistor gets close to turning off, then the gain will be lower. With one transistor off, the gain is zero. So keep in mind this circuit is only linear when input voltage is quite small. If you want to use it as a comparator, this is not an issue.