Here is the math that leads to that efficiency formula:

Define efficiency $\eta$ to be the ratio of the power going to accelerate the payload, $P_{pld}$, to the power in the payload plus the power in the exhaust,$P_e$, **as seen from the stationary observer**.

$P_{pld} = Fv = \dot{m} v_e v$, where $v$ is velocity of payload, $\dot{m}$ is the rate of exhaust mass (assumed positive), and $v_e$ is exhaust speed.

$P_e = \frac{\dot{m}}{2}(v_e-v)^2$

So,
$$ \eta = \frac{\dot{m}v_ev}{\dot{m} v_e v + \frac{\dot{m}}{2}(v_e-v)^2}=\frac{2\frac{v}{v_e}}{1+(\frac{v}{v_e})^2}$$.

Note that the change of momentum or thrust $\dot{m}v_e$ is independent of $v$ but the energy change per unit time of that exhaust, $(v_e - v)^2$, does depend on $v$. 

Some problems to note are that efficiency is 0 at $v = 0$.  Also, note that efficiency is 100% at $v = v_e$ but it really isn't because you are accelerating fuel that is being lost.  The change of energy per unit time in the load is actually $\frac{dE}{dt} = Fv - \frac{\dot{m}v^2}{2}$, but it was not counted because, well, it gets dropped.  But, this formula is a quick and dirty way to see that efficiency is best somewhere around $v = v_e$.

I prefer a different way to find the exhaust velocity that minimizes the fuel required to attain $\Delta v = v$. Use the rocket equation to get the total energy expended by fuel to attain the velocity change.

$$E = \frac{m_i - m_f}{2}v_e^2 = \frac{m_f}{2}(e^{v/v_e} - 1)v_e^2$$

The minimum energy per payload mass required is where the derivative goes to zero.
$$
\frac{1}{m_f}\frac{dE}{dv_e} = \frac{1}{2}[(2v_e - v)e^{v/v_e} - 2v_e] = 0.
$$
The minimum occurs at $v_e = 0.6275 v$, where $E = 1.54\frac{m_fv^2}{2}$.