Simplify $\frac {2^{n(1-n)}\cdot 2^{n-1}\cdot 4^n}{2\cdot 2^n\cdot 2^{(n-1)}}$ Simplify::
$$\frac {2^{n(1-n)}\cdot 2^{n-1}\cdot 4^n}{2\cdot 2^n\cdot 2^{(n-1)}}$$
My Attempt:
\begin{align}
&\frac {2^{n-n^2}\cdot 2^{n-1}\cdot 2^{2n}}{2\cdot 2^n\cdot 2^{n-1}}\\
&=\frac {2^{n-n^2+n-1+2n}}{2^{1+n+n-1}} \\
&=\frac {2^{4n-n^2-1}}{2^{2n}}
\end{align}
I could not move on. Please help me to continue.
 A: $$\frac {2^{4n-n^2-1}}{2^{2n}}=2^{4n-n^2-1-2n}=2^{2n-n^2-1}=2^{-(n-1)^2}$$
where I have used $\frac{a^b}{a^c}=a^{b-c}$
A: Try taking out $2^n$ and $2^{n-1}$ from the denominator and the numerator. I got $$\frac{1}{2^{(n-1)^2}}.$$
If you can't get it let me know. I'll show the work.
A: Since that is a quotient you can rewrite it as $$2^{(4n-n^2-1)-2n}= 2^{-(n-1)^2}$$
Notice indeed that $$(a-b)^2=a^2-2ab+b^2$$ so that you have $$(n-1)^2 = n^2-2n+1$$ so clearly also $$-(n-1)^2 = -n^2+2n-1$$ and by rewriting the middle term you get $$-(n-1)^2 = -n^2+4n-2n-1$$ because clearly $4n-2n =2n$ and as you can probably see this is the same as $$-(n-1)^2 = (-n^2+4n-1)-2n$$ by commutativity of addition
A: $\frac {2^{n(1-n)}.2^{n-1}.4^n}{2.2^n.2^{(n-1)}}$
First of all both N & D contains $2^{n-1}$ Eliminate them. Now you have, 
$\frac {2^{n(1-n)}.4^n}{2.2^n}$
$\frac {2^n.2^{-n^2}.2^{2n}}{2.2^n}$
Now both N and D have $2^n$ eliminate it.
$\frac {2^{-n^2}.2^{2n}}{2}$
= $\frac {2^{2n}}{2.2^{n^2}}$

Edit -

According to your answer -
=$\frac {2^{4n-n^2-1}}{2^{2n}}$
=$2^{4n-n^2-1-2n}$
=$2^{2n-n^2-1}$
=$2^{-(n^2-2n+1)}$
=$2^{-(n-1)^2}$
