How to find determinant of this matrix? Is there a manual method to find  $\det\left(XY^{-1}\right)$ ?
Let $$X=\left[ {\begin{array}{cc} 1 & 2 & 2^2 & \cdots & 2^{2012} \\ 
1 & 3 & 3^2 & \cdots & 3^{2012} \\
1 & 4 & 4^2 & \cdots & 4^{2012} \\  
\vdots  & \vdots  & \vdots  & \cdots & \vdots  \\ 
1 & 2014 & 2014^2 & \cdots & 2014^{2012} \\ \end{array} } \right], $$
$$Y=\left[ {\begin{array}{cc}\frac{2^2}{4} & \frac{3^2}{5} & \dfrac{4^2}{6} & \cdots & \dfrac{2014^2}{2016} \\ 
2 & 3 & 4 & \cdots & 2014 \\
2^2 & 3^2 & 4^2 & \cdots & 2014^{2} \\  
\vdots  & \vdots  & \vdots  & \cdots & \vdots  \\ 
2^{2012} & 3^{2012} & 4^{2012} & \cdots & 2014^{2012} \\ \end{array} } \right] $$.
Thanks in advance.

Not a solution, but a possible step in the right direction (too long for a comment):
Following Cramer's rule, we note that the solution $\vec y = (y_1,\dots,y_{2014})^T$ to 
$$X \,\vec y = \pmatrix{
\frac{2^2}{4} & \frac{3^2}{5} & \dfrac{4^2}{6} & \cdots & \dfrac{2014^2}{2016}
}^T$$
Will satisfy
$$
y_1 = \det(Y^T)/\det(X) = [\det(XY^{-1})]^{-1}
$$
With that in mind: if there is another way to solve this equation, it will probably be easier than actually computing the determinant.
Note also that $X$ is a Vandermonde matrix.
