<s>[INST] 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.
 [/INST] So, we've got $\dfrac{1}{-2^n \cdot -2^n} + \dfrac{1}{2^n \cdot 2^n} = \dfrac{1}{2^n \cdot 2^n} + \dfrac{1}{2^n \cdot 2^n} = \dfrac{2}{2^{2n}}$ since adding two of the same things up is the same as multiplying by $2$. 
Alright, so you have $\frac{2}{2^{2n}}$. That's good work! Now use the property that $\frac{a}{b} = ab^{-1}$, since $\frac{1}{b} = b^{-1}$. Different notations, same thing. 
Now, that means you have $\frac{2}{2^{2n}} = 2\cdot 2^{-2n}$. We're almost there: remember that $a^b \cdot a^c = a^{b+c}$, this can be seen by writing $a^b \cdot a^c = \underbrace{a \cdot a \cdots  a}_{b \, \text{times}} \cdot \underbrace{a \cdot a \cdots a}_{c \, \text{times}} = \underbrace{a \cdot a \cdots a}_{(b+c) \, \text{times}} = a^{b+c}$. 
Applying this to your case with $a = 2$, $b = 1$, $c = -2n$ we have $2 \cdot 2^{-2n} = 2^{1 - 2n} = 2^{-2n + 1}$. 
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