Proof improvement for $(a^2+b^2)(c^2+d^2)(e^2+f^2)(g^2+h^2) = A^2 + B^2$ given $(a+ib)(c+id)(e+if)(g+ih) = A + iB$ If $(a+ib)(c+id)(e+if)(g+ih) = A + iB$, prove that $(a^2+b^2)(c^2+d^2)(e^2+f^2)(g^2+h^2) = A^2 + B^2$
My approach is pretty straightforward: 
$$(a+ib)(c+id)(e+if)(g+ih)$$
$$((ac-bd)+i(ad+bc))(e+if)(g+ih)$$
$$(x+iy)(e+if)(g+ih)$$
$$((xe-yf)+i(xf+ye))(g+ih)$$
$$(u+iv)(g+ih)$$
At this point we can see that
$$A = ug-vh$$
$$B = uh+vg$$
$$A^2 + B^2 = (ug-vh)^2+(uh+vg)^2$$
And now you simply recursively unfold the above to get the desired result.
I can't shake the feeling that i'm missing something though. Maybe it's some extra step in my solution or maybe there's a much simpler solution.

For complex numbers $z_1$ and $z_2$, we have $$|z_1z_2|=|z_1||z_2|$$
So,
$$(a+ib)(c+id)(e+if)(g+ih) = A + iB$$
$$|(a+ib)(c+id)(e+if)(g+ih)| = |A + iB|$$
$$|(a+ib)||(c+id)||(e+if)||(g+ih)| = |A + iB|$$
$$(\sqrt{a^2+b^2})(\sqrt{c^2+d^2})(\sqrt{e^2+f^2})(\sqrt{g^2+h^2}) = (\sqrt{A^2+B^2})$$
Thus
$$(a^2+b^2)(c^2+d^2)(e^2+f^2)(g^2+h^2) = (A^2+B^2)$$
