Volume of $n$ dimensional ellipsoid

This comes out of substituting $\frac {x_i}{c_i}=y_i$ and using the volume of an $n$ dimensional ball in $n$ space plus the Jacobian transformation of volumes.

You know the formula for a ball. Note that the ellipsoid is just the ball scaled by a linear transformation. This linear transformation is diagonal with elements $c_1,c_2,\dots, c_n$. It is a standard fact (proved in Folland's Real Analysis, for example), that if $V$ is a Lebesgue measurable set and $L$ is a linear transformation, then the measure of $L(V)$ is $m(V)\cdot \det(L)$. It is easy to see that the determinant of the linear transformation of interest is $\prod_1^n c_i$, and this gives your desired result.

1) confirm the result in 2-D case

2) Assume the formula holds for n-D, then calculate the volumne in (n+1)-D using the techniques in the proof part of your wiki link, except for one small trick.

Suppose your ellipsoid in (n+1)-D has the axis $c_1,c_2,...c_{n+1}$. Then the n-D ellipsoid with $x_{n+1}=a$ is $$ \left(x_1,x_2,...x_n|\sum_{i=1}^n \frac{x_i^2}{c_i^2}<1-\frac{a^2}{c_{n+1}^2} \right) $$ the same as $$ \left(x_1,x_2,...x_n|\sum_{i=1}^n \frac{x_i^2}{d_i^2}<1 \right) $$ where $d_i=c_i\sqrt{1-a^2/c_{n+1}^2}$

3) So it holds for arbitrary dimension