Simple! But is it really?

It must be known that $\tau =2\pi$ , and thus $e^{ \tau i} = e^{2\pi i}$ . As well as Euler's formula: $e^{x i} = cos(x) + i sin(x)$
Using Euler's formula to compute $e^{2 \pi i}-e^{- \pi i}$ :
$e^{2 \pi i}-e^{- \pi i} = (cos( 2 \pi ) +i sin( 2 \pi ))-(cos(- \pi ) +i sin( - \pi )) =(1)-(-1) =2$
Thus $p^{3} + \frac{e^{2 \pi i}-e^{- \pi i}}{2} = p^{3} + \frac{2}{2} = p^{3} +1$ The only solution to $p^{3} +1 = q^{2}$ where both $p$ and $q$ are integers is $p=2$ and $q=3$ ,
It follows then that $q^{p} = 3^{2} = \textbf{9}$

I confess, I complicated the problem for no reason by bringing in $e$ , $\tau$ , $\pi$ & $i$ .

We know that $(e^ {\tau i}) = (e^ { 2\pi i}) = (1)$ and $(e^{-i \pi}) = (-1)$ . Following that, $(1)-(-1)=2$ and $\dfrac{2}{2} = 1$

So the simplified version would look something like :

$\huge p^3 + 1 = q^2$

And by Catalan's Conjecture, the only positive integer solutions are \text {q = 3 & p = 2 }