<s>[INST] 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.
 [/INST] $\frac {2^{n(1-n)}.2^{n-1}.4^n}{2.2^n.2^{(n-1)}}$
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 -
=$\frac {2^{4n-n^2-1}}{2^{2n}}$
=$2^{4n-n^2-1-2n}$
=$2^{2n-n^2-1}$
=$2^{-(n^2-2n+1)}$
=$2^{-(n-1)^2}$
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