(Perhaps this is something rather for MOverflow, don't know)
I've understood the concept of the Jordan-normal-form such that it is similar to the idea of diagonalization of a matrix, but can be applied even to non-diagonalizable matrices, such that for a diagonalizable matrix $A$ we can write $$ A = M \cdot D \cdot M^{-1} $$, or for convenience let's write for $M^{-1}$ the letter $W$ for now (which might be acceptable as a memorizable methapher for the reciprocal letter of $M$ ;-)).
So if diagonalization is possible, then $$ A = M \cdot D \cdot W \qquad \qquad \text{ D is diagonal } $$ and powers of A and the inverse of A (provided it is invertible) are expressible $$ A^p = M \cdot D^p \cdot W \\ A^{-1} = M \cdot D^{-1} \cdot W $$
If $A$ is not diagonalizable, then we can still get its Jordan normal form similarly like $$ A = M \cdot J \cdot W \qquad \qquad \text{ J is in Jordan form } $$ where $J$ contains triangular blocks, by convention arranged as upper triagular matrix (at least if we look at wolfram-alpha; here I use the transpose notation for readability).
Now I looked at the matrices of Stirling numbers second kind (S2) and of first kind (S1) which are inverses of each other. For $A=S_2$ I get $$ A = S_2 = \small \begin{bmatrix} 1 & . & . & . & . & . \\ 1 & 1 & . & . & . & . \\ 1 & 3 & 1 & . & . & . \\ 1 & 7 & 6 & 1 & . & . \\ 1 & 15 & 25 & 10 & 1 & . \\ 1 & 31 & 90 & 65 & 15 & 1 \end{bmatrix}$$ $$ M,J = \small \begin{array} {|rrrrrr|rrrrrr|} 1 & . & . & . & . & . & 1 & . & . & . & . & . \\ . & 1 & . & . & . & . & 1 & 1 & . & . & . & . \\ . & 1 & 3 & . & . & . & . & 1 & 1 & . & . & . \\ . & 1 & 13 & 18 & . & . & . & . & 1 & 1 & . & . \\ . & 1 & 50 & 205 & 180 & . & . & . & . & 1 & 1 & . \\ . & 1 & 201 & 1865 & 4245 & 2700 & . & . & . & . & 1 & 1 \end{array} $$
and for $A=S_1=S_2^{-1}$ $$ A = S_1 = \small \begin{bmatrix} 1 & . & . & . & . & . \\ -1 & 1 & . & . & . & . \\ 2 & -3 & 1 & . & . & . \\ -6 & 11 & -6 & 1 & . & . \\ 24 & -50 & 35 & -10 & 1 & . \\ -120 & 274 & -225 & 85 & -15 & 1 \end{bmatrix} $$ I get $$ M,J = \small \begin{array} {|rrrrrr|rrrrrr|} 1 & . & . & . & . & . & 1 & . & . & . & . & . \\ . & -1 & . & . & . & . & 1 & 1 & . & . & . & . \\ . & 2 & 3 & . & . & . & . & 1 & 1 & . & . & . \\ . & -6 & -23 & -18 & . & . & . & . & 1 & 1 & . & . \\ . & 24 & 180 & 335 & 180 & . & . & . & . & 1 & 1 & . \\ . & -120 & -1594 & -5330 & -6555 & -2700 & . & . & . & . & 1 & 1 \end{array} $$ Here the inversion is not on $J$ but on $M$ which is not wanted. However, if I express $S_1$ via the inversion-formula similar to that of the diagonalization I keep the "correct" version of $M$ but the matrix $J^{-1}$ has a form which I thought would not be acceptable as Jordan normal form; I get: $$J^{-1}= \small \begin{bmatrix} 1 & . & . & . & . & . \\ -1 & 1 & . & . & . & . \\ 1 & -1 & 1 & . & . & . \\ -1 & 1 & -1 & 1 & . & . \\ 1 & -1 & 1 & -1 & 1 & . \\ -1 & 1 & -1 & 1 & -1 & 1 \end{bmatrix}$$
Q: Is one of the representations of $S_1$ invalid or are they both acceptable as Jordan normal forms?
Q2: And which kind of representation should I prefer in the expectation that I want to use that concept for infinite matrices later?