Let's set $c=1$ for simplicity.

Using your observations, it suffices  to show that (just combine the second and third equations you write down)
$$
  \dot \gamma = \vec u \cdot \frac{d}{dt}(\gamma \vec u)
$$
To prove this, the following facts are useful:
$$
  \dot \gamma = \gamma^3\vec u \cdot\dot{\vec u}, \qquad \gamma^2\vec u^2 +1 = \gamma^2 
$$
Now just compute
\begin{align}
\vec u \cdot \frac{d}{dt}(\gamma \vec u)
  &=\vec u \cdot (\dot \gamma \vec u + \gamma \dot{\vec u}) \\
&=\vec u \cdot (\gamma^3(\vec u \cdot \dot{\vec u})\vec u + \gamma \dot{\vec u}) \\
&= \gamma \vec u \cdot \dot{\vec u}(\gamma^2 \vec u^2 + 1) \\
&= \gamma^3\vec u \cdot\dot {\vec u} \\
&= \dot \gamma
\end{align}