Consider the table containing the Stirling numbers of the second kind $S\left( n,k\right) $. From this table construct a square matrix $M$ containing $N$ differents rows from the table ($n_{1},\ldots,n_{N}$), and colums $1$ up to $N$. So:
$$M = \begin{bmatrix} S\left( n_{1},1\right) & \ldots & S\left( n_{1},N\right) \\ \vdots & & \vdots\\ S\left( n_{N},1\right) & \ldots & S\left( n_{N},N\right) \end{bmatrix}$$
where $n_{i}\in %TCIMACRO{\U{2115} } %BeginExpansion \mathbb{N} %EndExpansion _{0}$ and $n_{1}<n_{2}<\ldots<n_{N}$. Since $S\left( n,1\right)=1$ if $n>0$ $M$ is equal to:
$$M = \begin{bmatrix} 1 & \ldots & S\left( n_{1},N\right) \\ \vdots & & \vdots\\ 1 & \ldots & S\left( n_{N},N\right) \end{bmatrix}$$
An example of $M$ where $N=3$ and $n=\left\{ 1,3,5\right\} $:
$$M_{ex} = \begin{bmatrix} 1 & 0 & 0\\ 1 & 3 & 1\\ 1 & 15 & 25 \end{bmatrix}$$
where $\det\left( M_{ex}\right) =60$.
My question is the following: Is there a known result that states that the determinant of $M$ is always non-zero. If not, how would one go about in showing this?
I suspect that one might somehow use either the explicit formula $$S\left( n,k\right) =\frac{1}{k!}\sum \limits_{j=0}^{k} \left( -1\right) ^{k-j}\binom{k}{j}j^{n}$$ or one of the (many) recursive relations $$S\left( n+1,k\right)=kS\left( n,k\right) +S\left( n,k-1\right)$$ $$S\left( n+1,k+1\right)=\sum \limits_{j=k}^{n} \binom{n}{j} = S\left( j,k\right)$$
but I fail to see how.
Really appreciate any suggestions to solve this problem.
Thanks in advance.