Getting from $E^2 - p^2c^2 = m^2c^4$ to $E = \gamma mc^2$ What is each mathematical step (in detail) that one would take to get from:
$E^2 - p^2c^2 = m^2c^4$ 
to
$E = \gamma mc^2$,
where $\gamma$ is the relativistic dilation factor.
This is for an object in motion.
NOTE: in the answer, I would like full explanation. E.g. when explaining how to derive $x$ from $\frac{x+2}{2}=4$, rather than giving an answer of "$\frac{x+2}{2}=4$, $x+2 = 8$, $x = 6$" give one where you describe each step, like "times 2 both sides, -2 both sides" but of course still with the numbers on display. (You'd be surprised at how people would assume not to describe in this detail).

Starting with relativistic momentum
$$p^2 = \left( \gamma m v \right)^2 = \frac{m^2 v^2}{1 - \frac{v^2}{c^2}}$$
one than gets
$$E =  \pm \sqrt{ m^2 c^4 + p^2 c^2 } = \pm \sqrt{ m^2 c^4 + \frac{m^2 v^2 c^2}{1 - \frac{v^2}{c^2}} } = \pm mc^2 \sqrt{\frac{1- \frac{v^2}{c^2}}{1- \frac{v^2}{c^2}} + \frac{\frac{v^2}{c^2}}{1- \frac{v^2}{c^2}}}  = \pm \gamma mc^2$$
