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?
 A: 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)
A: This is due to the fact that $(-1)^{n+2}$ is the same thing as $(-1)^n$.  Hence, $(-1)^n2^{n+2}=(-1)^{n+2}2^{n+2}$, and by the properties of exponents, this is equal to $(-2)^{n+2}$.
A: 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).
