Theorem: Inverse of Hilbert Matrix

Let $H_n$ be the Hilbert matrix of order $n$:
:$\begin{bmatrix} a_{i j} \end{bmatrix} = \begin{bmatrix} \dfrac 1 {i + j - 1} \end{bmatrix}$
Then its inverse $H_n^{-1} = \sqbrk b_n$ can be specified as:
:$\begin{bmatrix} b_{i j} \end{bmatrix} = \begin{bmatrix} \dfrac {\paren {-1}^{i + j} \paren {i + n - 1}! \paren {j + n - 1}!} {\paren {\paren {i - 1}!}^2 \paren {\paren {j - 1}!}^2 \paren {n - j}! \paren {n - i}! \paren {i + j - 1} } \end{bmatrix}$

Proof:
From Hilbert Matrix is Cauchy Matrix, $H_n$ is a special case of a Cauchy matrix:
:$\begin{bmatrix} c_{i j} \end{bmatrix} = \begin{bmatrix} \dfrac 1 {x_i + y_j} \end{bmatrix}$
where:
:$x_i = i$
:$y_j = j - 1$
From Inverse of Cauchy Matrix, the inverse of the square Cauchy matrix of order $n$ is:
:$\begin{bmatrix} b_{i j} \end{bmatrix} = \begin{bmatrix} \dfrac {\ds \prod_{k \mathop = 1}^n \paren {x_j + y_k} \paren {x_k + y_i} } {\ds \paren {x_j + y_i} \paren {\prod_{\substack {1 \mathop \le k \mathop \le n \\ k \mathop \ne j} } \paren {x_j - x_k} } \paren {\prod_{\substack {1 \mathop \le k \mathop \le n \\ k \mathop \ne i} } \paren {y_i - x_k} } } \end{bmatrix}$
Thus $H_n^{-1}$ can be specified as:
:$\begin{bmatrix} b_{i j} \end{bmatrix} = \begin{bmatrix} \dfrac {\ds \prod_{k \mathop = 1}^n \paren {i + k - 1} \paren {j + k - 1} } {\ds \paren {i + j - 1} \paren {\prod_{\substack {1 \mathop \le k \mathop \le n \\ k \mathop \ne i} } \paren {i - k} } \paren {\prod_{\substack {1 \mathop \le k \mathop \le n \\ k \mathop \ne j} } \paren {j - k} } } \end{bmatrix}$
First, from Product of Products:
:$\ds \prod_{k \mathop = 1}^n \paren {i + k - 1} \paren {j + k - 1} = \prod_{k \mathop = 1}^n \paren {i + k - 1} \prod_{k \mathop = 1}^n \paren {j + k - 1}$
We address in turn the various factors of this expression for $b_{i j}$.
{{begin-eqn}}
{{eqn | l = \prod_{k \mathop = 1}^n \paren {i + k - 1}
      | r = \prod_{k \mathop = 0}^{n - 1} \paren {i + k}
      | c = Translation of Index Variable of Product
}}
{{eqn | r = i^{\overline n}
      | c = {{Defof|Rising Factorial}}
}}
{{eqn | r = \frac {\paren {i + n - 1}!} {\paren {i - 1}!}
      | c = Rising Factorial as Quotient of Factorials
}}
{{end-eqn}}
and similarly:
:$\ds \prod_{k \mathop = 1}^n \paren {j + k - 1} = \frac {\paren {j + n - 1}!} {\paren {j - 1}!}$
Then:
{{begin-eqn}}
{{eqn | l = \prod_{\substack {1 \mathop \le k \mathop \le n \\ k \mathop \ne i} } \paren {i - k}
      | r = \paren {\prod_{1 \mathop \le k \mathop < i} \paren {i - k} } \paren {\prod_{i \mathop < k \mathop \le n} \paren {i - k} }
      | c = 
}}
{{eqn | r = \paren {i - 1}! \paren {\prod_{i \mathop < k \mathop \le n} \paren {i - k} }
      | c = {{Defof|Factorial}}
}}
{{eqn | r = \paren {i - 1}! \paren {\prod_{0 \mathop < k \mathop \le n - i} \paren {-k} }
      | c = Translation of Index Variable of Product
}}
{{eqn | r = \paren {i - 1}! \paren {-1}^{n - i} \paren {\prod_{0 \mathop < k \mathop \le n - i} k}
      | c = 
}}
{{eqn | r = \paren {i - 1}! \paren {-1}^{n - i} \paren {n - i}!
      | c = {{Defof|Factorial}}
}}
{{end-eqn}}
and similarly:
:$\ds \prod_{\substack {1 \mathop \le k \mathop \le n \\ k \mathop \ne j} } \paren {j - k} = \paren {j - 1}! \paren {-1}^{n - j} \paren {n - j}!$
Thus we can write:
{{begin-eqn}}
{{eqn | l = \begin{bmatrix} b_{i j} \end{bmatrix}
      | r = \frac {\paren {\dfrac {\paren {i + n - 1}!} {\paren {i - 1}!} } \paren {\dfrac {\paren {j + n - 1}!} {\paren {j - 1}!} } } {\paren {i + j - 1} \paren {i - 1}! \paren {-1}^{n - i} \paren {n - i}! \paren {j - 1}! \paren {-1}^{n - j} \paren {n - j}!}
      | c = 
}}
{{eqn | r = \frac {\paren {-1}^{i + j} \paren {i + n - 1}! \paren {j + n - 1}!} {\paren {\paren {i - 1}!}^2 \paren {\paren {j - 1}!}^2 \paren {n - i}! \paren {n - j}! \paren {i + j - 1} }
      | c = 
}}
{{end-eqn}}
{{qed}}