We will begin by proving that \(\overline{z_1 z_2} = \overline{z_1} \; \overline{z_2} \) for any two complex numbers \(z_1\) and \(z_2\)
Let \(z_1 = x_1 + iy_1\) and \(z_2 = x_2 + i y_2\). Then \begin{align*} \overline{z_1 z_2} &= \overline{(x_1 + iy_1)(x_2 + i y_2)}\\[5pt] &= \overline{x_1x_2 + i(x_1y_2 + y_1x_2)-y_1y_2} \\[5pt] &= x_1y_2 - i(x_1y_2 + y_1x_2) -y_1y_2\\[5pt] &= (x_1 - iy_1)(x_2 - iy_2)\\[5pt] &= \overline{(x_1 + iy_1)} \;\overline{(x_2 + iy_2)}\\[5pt] &= \overline{z_1} \; \overline{z_2} \end{align*}
The rest of the proof follows fairly simply. I will try to be as explicit as possible. We can write \[ \overline{z^4} = \overline{z^2 z^2} =\overline{z^2}\;\overline{z^2} = (\overline{z}\;\overline{z})(\overline{z}\;\overline{z})= \overline{z}\;^4\]
Q.E.D.