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 $s \left({0, 0}\right)$.


 * $\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}$

Thenconsider Stirling's triangle of the second kind expressed as a (square) matrix $\mathbf B$, with the top left element holding $\displaystyle \left \{ {0 \atop 0}\right\}$.


 * $\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:
 * $\displaystyle \left[{n \atop m}\right] = \left({-1}\right)^{n + m} s \left({n, m}\right)$

From First Inversion Formula for Stirling Numbers:
 * $\displaystyle \sum_k \left[{n \atop k}\right] \left\{ {k \atop m}\right\} \left({-1}\right)^{n - k} = \delta_{m n}$

From Second Inversion Formula for Stirling Numbers:
 * $\displaystyle \sum_k \left\{ {n \atop k}\right\} \left[{k \atop m}\right] \left({-1}\right)^{n - k} = \delta_{m n}$

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.