How do you simplify $3^{\frac{(-1)^n + 1}{2}}$ I am solving non linear recursive relations and I stumbled upon this: $$x(n)*x(n+1)=3,\ x(0)=3$$
If you start calculating the values after $0$ you will notice that a pattern emerges: $$x(0) = 3$$ $$x(1) = 1$$ $$x(2) = 3$$ $$x(3) = 1$$ $$...$$
From this I concluded that $x(n) = (-1)^n + 2$
BUT! If you try to solve the problem more conservatively you will convert the relation to a linear one: 
$log_{3} (x(n)) + log_{3}(x(n+1)) = 1$ $Set\ g(n)=log_{3}(x(n)),\ g(0) = log_{3}(x(0))=1$
$g(n+1) + g(n) = 1 \Rightarrow g(n)=\frac{1}{2}*(-1)^n + \frac{1}{2} \Rightarrow x(n) = 3^{\frac{1}{2}*((-1)^n + 1)} $
The fact that $3^{\frac{1}{2}*((-1)^n + 1)} = (-1)^n + 2$ can also be confirmed by graphing the two functions. My question is: is there a way to go from the one to the other instead of just proving that they are equal?

This is of the same type of situation as saying that $2$ and $1 + 1$ are the same. They are different expressions for the same value.
In your case, let $f\left(n\right) = \left(-1\right)^n + 2$, for $n \ge 0 \text{ and } n \in N$. For even values of $n$, the value is $1 + 2 = 3$ and for odd values of $n$, the value is $-1 + 3 = 1$.
Similarly, let $h\left(n\right) = 3^{\frac{\left(-1\right)^n + 1}{2}}$. For $n$ being even here, the value of $\left(-1\right)^n = 1$, so the function becomes $3^{\frac{\left(1 + 1 \right)}{2}} = 3^1 = 3$. With odd values of $n$, $\left(-1\right)^n = -1$, so the function becomes $3^{\frac{\left(-1 + 1\right)}{2}} = 3^0 = 1$.
As you can see, the $2$ functions always have the same value for all non-negative values of $n$. They are just different ways to express the set of results.  However, I prefer $f\left(n\right) = \left(-1\right)^n + 2$ as, to me at least, it's simpler to understand and use.
