As pointed out by Richard Stanley in the comments, from the generating function of the Bell polynomials one finds $$N(n,k) = \frac{1}{k!}\sum_{i=1}^k(-1)^{k-i}\binom ki (ix) _n.$$ Now use the fact that $$(ix) _n=\sum _{j=0}^n s(n,j)i^j x^j,$$ and that $$\frac{1}{k!}\sum _{i=0}^k (-1)^{k-i}\binom{k}{i} i^j=S(j,k),$$ to obtain $$N(n,k)=\sum _{j\geq 1} s(n,j)x^jS(j,k).$$ In other words, $N=sDS$, where $s,S$ are the lower triangular matrices of Stirling numbers of first and second kind, respectively. To finish off, notice that $s^{-1}=S$ by Stirling number duality.

As pointed out by Richard Stanley in the comments, from the generating function of the Bell polynomials one finds $$N(n,k) = \frac{1}{k!}\sum_{i=1}^k(-1)^{k-i}\binom ki (ix) _n.$$ Now use the fact that $$(ix) _n=\sum _{j=0}^n s(n,j)i^j x^j,$$ and that $$\frac{1}{k!}\sum _{i=0}^k (-1)^{k-i}\binom{k}{i} i^j=S(j,k),$$ to obtain $$N(n,k)=\sum _{j=k}^n s(n,j)x^jS(j,k).$$ In other words, $N=sDS$, where $s,S$ are the lower triangular matrices of Stirling numbers of first and second kind, respectively. To finish off, notice that $s^{-1}=S$ by Stirling number duality.

As pointed out by Richard Stanley in the comments, from the generating function of the Bell polynomials one finds $$N(n,k) = \frac{1}{k!}\sum_{i=1}^k(-1)^{k-i}\binom ki (ix) _n.$$ Now use the fact that $$(ix) _n=\sum _{j=0}^n s(n,j)i^j x^j,$$ and that $$\frac{1}{k!}\sum _{i=0}^k (-1)^{k-i}\binom{k}{i} i^j=S(j,k),$$ we obtain $$N(n,k)=\sum _{j\geq 1} s(n,j)x^jS(j,k).$$ In other words, $N=sDS$, where $s,S$ are the lower triangular matrices of Stirling numbers of first and second kind, respectively. To finish off, notice that $s^{-1}=S$ by Stirling number duality.

As pointed out by Richard Stanley in the comments, from the generating function of the Bell polynomials one finds $$N(n,k) = \frac{1}{k!}\sum_{i=1}^k(-1)^{k-i}\binom ki (ix) _n.$$ Now use the fact that $$(ix) _n=\sum _{j=0}^n s(n,j)i^j x^j,$$ and that $$\frac{1}{k!}\sum _{i=0}^k (-1)^{k-i}\binom{k}{i} i^j=S(j,k),$$ to obtain $$N(n,k)=\sum _{j\geq 1} s(n,j)x^jS(j,k).$$ In other words, $N=sDS$, where $s,S$ are the lower triangular matrices of Stirling numbers of first and second kind, respectively. To finish off, notice that $s^{-1}=S$ by Stirling number duality.