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?

The exponent rules (for positive integer exponents, at any rate) are:


*

*$(a^n)^m = a^{nm}$

*$(ab)^n = a^nb^n$

*$a^na^m = a^{n+m}$.


Here, $a$ and $b$ are any real numbers, and $n$ and $m$ are positive integers. (The rules are valid in greater generality, but one has to be careful with the values of $a$ and $b$; also, the 'explanation' below is not valid for exponents that are not positive integers.)
To see these, remember what the symbols mean: $a^1 = a$, and $a^{n+1}=a^na$; that is, $a^k$ "means"
$$a^k = \underbrace{a\times a\times\cdots\times a}_{k\text{ factors}}$$
The following can be proven formally with induction, but informally we have:
$$\begin{align*}
(a^n)^m &= \underbrace{a^n\times a^n\times\cdots\times a^n}_{m\text{ factors}}\\
&= \underbrace{\underbrace{a\times\cdots\times a}_{n\text{ factors}}\times\cdots \times \underbrace{a\times\cdots\times a}_{n\text{ factors}}}_{m\text{ products}}\\
&= \underbrace{a\times\cdots \times a\times a\times\cdots \times a\times\cdots \times a}_{nm\text{ factors}}\\
&= a^{nm}
\end{align*}$$
Similarly,
$$\begin{align*}
(ab)^n &= \underbrace{(ab)\times (ab)\times\cdots\times (ab)}_{n\text{ factors}}\\
&= \underbrace{(a\times a\times\cdots \times a)}_{n\text{ factors}}\times\underbrace{(b\times b\times\cdots \times b)}_{n\text{ factors}}\\
&= a^nb^n,
\end{align*}$$
and
$$\begin{align*}
a^{n+m} &= \underbrace{a\times a\times\cdots\times a}_{n+m\text{ factors}}\\
&= \underbrace{(a\times a\times \cdots \times a)}_{n\text{ factors}}\times\underbrace{(a\times a\times \cdots \times a)}_{m\text{ factors}}\\
&= a^na^m.
\end{align*}$$
You have
$$(-1)^n2^{n+2}.$$
Because the bases are different ($-1$ and $2$), you do not apply rule 3 above (which is what you seem to want to do); instead, you want to try to apply rule 2. You can't do that directly because the exponents are different. However, since $(-1)^2 = (-1)(-1) = 1$, we can first do this:
$$(-1)^n2^{n+2} = 1(-1)^n2^{n+2} = (-1)^2(-1)^n2^{n+2};$$
then we apply rule 3 to $(-1)^2(-1)^n$ to get $(-1)^{2+n} = (-1)^{n+2}$, and now we have the situation of rule 2, so we get:
$$(-1)^n2^{n+2} = (-1)^{n+2}2^{n+2} = \bigl( (-1)2\bigr)^{n+2} = (-2)^{n+2}.$$
(You seem to be trying to apply a weird combination of rules 2 and 3, to get that $a^nb^m = (ab)^{n+m}$; this is almost always false; the exponent rules don't let you do that)
