Prove that the product of the $n$ roots of $e^{-i\cdot \frac{\pi }{2}}$ is given by $e^{i\cdot \frac{\left(2n-3\right)\pi }{2}}$ Let $w=e^{-i\cdot \frac{\pi }{2}}$. 

Here's how to tried to solve this:

$\sqrt[n]{w}=e^{i\cdot \left(-\frac{\pi }{2n}+\frac{2\pi }{n}k\right)},\:k\in \left(0;\:1;\:2;\:...;\:n-1\right)$ 

Factoring out the $1/n$, we get $e^{\frac{i}{n}\cdot \left(-\frac{\pi }{2}+2\pi k\right)}$.

Since for any value of $k$ we get , $e^{\frac{i}{n}\cdot \left(-\frac{\pi }{2}\right)}$, that is, all the roots are the same, the product of the n-th roots would be $\left(e^{\frac{i}{n}\cdot \:\left(-\frac{\pi \:}{2}\right)}\right)^{\left(n-1\right)}$. I get stuck here.

How do I get to $e^{i\cdot \frac{\left(2n-3\right)\pi }{2}}$?
Your approach seems correct, except for a small mistake in the "product". I have solved it as follows.

The $n$ roots of $e^{- \iota \frac{\pi}{2}}$ are given by $e^{\iota \left( - \frac{\pi}{2n} + \frac{2 \pi k}{n} \right)}$ for $k = 0, 1, 2, \cdots, n - 1$

\begin{align*}
\prod\limits_{k = 0}^{n - 1} e^{\iota \left( - \frac{\pi}{2n} + \frac{2 \pi k}{n} \right)} &amp;= e^{- \iota \frac{\pi}{2n}} \prod\limits_{k = 1}^{n - 1} e^{\iota \left( - \frac{\pi}{2n} + \frac{2 \pi k}{n} \right)} \\ 
&amp;= e^{- \iota \frac{\pi}{2n}} \cdot e^{\sum\limits_{k = 1}^{n} \iota \left( - \frac{\pi}{2n} + \frac{2 \pi k}{n} \right)} \\ 
&amp;= e^{- \iota \frac{\pi}{2}} \cdot e^{\iota \left( - \frac{\pi}{2n} \left( n - 1 \right) + \frac{2\pi}{n} \frac{n \left( n - 1 \right)}{2} \right)} \\ 
&amp;= e^{- \iota \left( \frac{\pi}{2n} \left( 1 + n - 1 - 2 n \left( n - 1 \right) \right) \right)} \\ 
&amp;= e^{\iota \frac{\pi}{2} \left( 2n - 3 \right)}\end{align*}
Hint

$x^n=i^{-1}=-i$

$\implies x^n+i=0$

Vieta's formula says the product $$=(-1)^ni=(e^{i\pi})^n\cdot e^{i\pi/2}$$