linear algebra - Finding the closed of form of the determinant of the Hilbert matrix

In my studies of matrix theory I came across the famous Hilbert matrix, which is a square $n \times n$ matrix $H$ with entries given by: $h_{ij} = \frac{1}{i+j-1}$ and this is an example of a Cauchy matrix, which is a matrix $C_n$ of the form $c_{ij} = \frac{1}{x_i+y_j}$ and for this matrix there is the well known formula for the determinant:

$det(C) = \dfrac {\displaystyle \prod_{1 \mathop \le i \mathop < j \mathop \le n} \left({x_j - x_i}\right) \left({y_j - y_i}\right)} {\displaystyle \prod_{1 \mathop \le i, \, j \mathop \le n} \left({x_i + y_j}\right)}$

Now I think I can substitute the sequences for the Hilbert matrix but I cannot see how to get the closed form they got here (under Properties):

$det(H) = \frac{c_n^4}{c_{2n}}$ where $c_n = \prod_{i=1}^{n-1} i^{n-1} =\prod_{i=1}^{n-1} i!$

and I was hoping someone would please help me obtain the closed form. Thanks.