Multiplying exponents with variables inside Why is 
$$(-1)^n(2^{n+2}) = (-2)^{n+2} ?$$
My thinking is that $-1^n \times 2^{n+2}$ should be $-2^{2n+2}$ but clearly this is not the case. Why is the variable essentially ignored, is there a special case of multiplication I'm unaware of?

You seem to be misapplying the rules for simplifying exponents.
The relevant rules here are:
1) If you have $\color{maroon}a^x \cdot \color{maroon}a^y$, you may write $\color{maroon}a^x \cdot \color{maroon}a^y = \color{maroon}a^{x+y}$. So if the $\color{maroon}{\text{base}}$ is the same in a product of exponential terms, you add the exponents.
2) If you have $x^{\color{darkgreen}b} \cdot  y^{\color{darkgreen}b}$, you may write  $ x^{\color{darkgreen}b} \cdot\ y^{\color{darkgreen}b} =(xy)^{\color{darkgreen}b}$. So if the $\color{darkgreen}{\text{exponent}}$ is the same in a product of exponential terms, you multiply the bases.
You have neither of the two going on with your expression  $(-1)^n 2^{n+2}$; the bases are different and the exponents are different, so, you can not apply either of the above rules directly (and certainly not both as it seems you tried).
However, you can set things up so that the second rule can be used, as Hayden and Arturo did in their  answers.
Or you could do it this way:
$$\eqalign{
(-1)^n\color{maroon}{ 2^{n+2}}
&\quad\buildrel{ \color{maroon}{ \text{ rule 1)  }} }\over = 
\quad(-1)^n \color{maroon}{2^{n }\cdot2^2}\cr
&\quad\buildrel{ \phantom{\text{ rule 1)  }} }\over = \quad 
 \color{darkgreen}{(-1)^n 2^{n }}\cdot2^2 \cr
&\quad\buildrel{\color{darkgreen}{\text{rule 2)}}}\over = \quad
\color{darkgreen}{(-2)^{n }}\cdot2^2  \cr
&\quad \buildrel{ \phantom{\text{ rule 1)  }} }\over = 
 \quad{(-2)^{n }}(-2)^2 \cr
&\quad \buildrel{\text{rule 1)}}\over = \quad(-2)^{n+2}.
}
$$
Note in the next to last equality, we needed to use a "trick" in order to aplly rule 1).
