Inverse of Stirling's Triangle expressed as Matrix

Theorem
Consider Stirling's triangle of the first kind (signed) expressed as a (square) matrix $\mathbf A$, with the top left element holding $\map s {0, 0}$.


 * $\begin{pmatrix}

1 &    0 &       0 &      0 &      0 &     0 & \cdots \\ 0 &    1 &       0 &      0 &      0 &     0 & \cdots \\ 0 &   -1 &       1 &      0 &      0 &     0 & \cdots \\ 0 &    2 &      -3 &      1 &      0 &     0 & \cdots \\ 0 &   -6 &      11 &     -6 &      1 &     0 & \cdots \\ 0 &   24 &     -50 &     35 &    -10 &     1 & \cdots \\ \vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \ddots \\ \end{pmatrix}$

Then consider Stirling's triangle of the second kind expressed as a (square) matrix $\mathbf B$, with the top left element holding $\ds {0 \brace 0}$.


 * $\begin{pmatrix}

1 & 0 &  0 &    0 &    0 &    0 & \cdots \\ 0 & 1 &  0 &    0 &    0 &    0 & \cdots \\ 0 & 1 &  1 &    0 &    0 &    0 & \cdots \\ 0 & 1 &  3 &    1 &    0 &    0 & \cdots \\ 0 & 1 &  7 &    6 &    1 &    0 & \cdots \\ 0 & 1 & 15 &   25 &   10 &    1 & \cdots \\ \vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \ddots \\ \end{pmatrix}$

Then:
 * $\mathbf A = \mathbf B^{-1}$

that is:
 * $\mathbf B = \mathbf A^{-1}$

Proof
First note that from Relation between Signed and Unsigned Stirling Numbers of the First Kind:
 * $\ds {n \brack m} = \paren {-1}^{n + m} \map s {n, m}$

From First Inversion Formula for Stirling Numbers:
 * $\ds \sum_k {n \brack k} {k \brace m} \paren {-1}^{n - k} = \delta_{m n}$

From Second Inversion Formula for Stirling Numbers:
 * $\ds \sum_k {n \brace k} {k \brack m} \paren {-1}^{n - k} = \delta_{m n}$

By definition of matrix multiplication, the element $a_{r n}$ of the matrix formed by multiplying the two matrices above.

As can be seen, this results in the identity matrix.