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.

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
