I often see this diagram of a current mirror (as shown below).
As far as I know, the purpose of a current mirror is the ensure that the collector current for both transistors are equal.
This can simply be achieved by making sure that their base-emitter voltage is the same. This can be done without shorting the base and collector of the left hand side transistor... Is shorting it redundant in any ways?
This can simply be achieved by making sure that their base-emitter voltage is the same.
Not quite. The collector current can be written as
$$I_C = \beta_0\left(1 + \frac{V_{CB}}{V_A}\right)I_B$$
and so depends on the base current and the collector-base voltage (Early effect). Connecting the collector and base together $(V_{CB}=0)$ removes this dependence on the collector-base voltage and the relationship simplifies to
$$I_C = \beta_0 I_B$$
Since $V_{BE1} = V_{BE2}$, the base currents are equal (assuming identical transistors etc.) and so we can then write
$$I_{REF} \equiv \frac{V_{DD} - V_{BE1}}{R_1} = I_{C1} + I_{B1} + I_{B2} = I_{B2}\left(2 + \beta_0\right)$$
and it follows that
$$I_{C2}=I_{REF}\frac{\beta_0}{2 + \beta_0}\left(1 + \frac{V_{CB2}}{V_A}\right) $$
If, on the other hand, $V_{CB1} \ne 0$ (for example, place a resistor between the collector and base of Q1 rather than a wire), the equation relating $I_{C2}$ to $I_{REF}$ is more complicated
$$I_{REF} \equiv \frac{V_{DD} - V_{BE1} - V_{CB1}}{R_1}$$
$$I_{C2}=I_{REF}\frac{\beta_0\left(1 + \frac{V_{CB2}}{V_A}\right)}{2 + \beta_0\left(1 + \frac{V_{CB1}}{V_A}\right)}$$
Shorting collector to base is NOT redundant, it serves the useful purpose of making conduction in the two transistors similar. In particular, the construction of a transistor includes a thin base region between emitter and collector, and that base region has an effective electrical resistance. Forcing all current in one transistor through the base creates a voltage drop in the base spreading resistance (often symbolized Rbb), while the (second) transistor takes only a small fraction of its emitter current through that resistance.
Ebers-Moll ideal transistor: $$I_c = \alpha I_{sat} \exp {q_e * V_{be} \over { k * T }}$$
and with base resistance $$I_c = \alpha I_{sat} \exp ({q_e * (V_{be} -I_b * R_{bb})\over { k * T }})$$
In the case of base-collector connection, Ib is a small fraction of the collector current. If you don't connect the collector of the leftmost transistor, though, Ib is identical to the emitter current (which means it is larger than the collector current). To match the two transistors' operation, you want the base-collector connected.
The open-collector use of B-E diode also changes the 'alpha' factor to exactly one, but that's less important.
