If $n$ is a positive integer, then $(-2^n)^{-2} + (2^{-n})^2 = 2^{-2n+1}$ I'm not sure why $$(-2^n)^{-2} + (2^{-n})^2=2^{-2n+1}$$
I have been going over this equation for a while now, noticing, and have successfully got quite far in the equation, finding that 
$$ (-2^n)^{-2} + (2^{-n})^2 \implies \frac{1}{(-2^n)^2} + \frac{1}{(2^n)^2} $$
which I think then becomes $ \dfrac{2}{2^{2n}}$
But then I get stuck.

$(-2^n)^{-2} + (2^{-n})^2=$
$\frac 1{(-2^n)^2} + (\frac{1}{2^n})^2 = $
$\frac 1{(2^n)^2} + \frac 1 {(2^n)^2} = $
$\frac 1{2^{2n}} + \frac 1{2^{2n}} = $
$\frac 2{2^{2n}} = $
$\frac 1{2^{2n-1}} = $
$2^{-(2n-1)} = $
$2^{-2n+1}$
Or if you are more comfortable with the rules of exponents:
$(-a)^{b_{\text{even}}} = a^b$ so
$(-2^n)^{-2} + (2^{-n})^2 = (2^n)^{-2} + (2^{-n})^2 = 2^{-2n} + 2^{-2n} = 2*2^{-2n} = 2^{-2n + 1}$
