(1)

\(2\leq n,0\leq k\leq n\)として、
\[ \left(-1\right)^{n-k}A\left(n,k\right)=\sum_{1\leq a_{1}<a_{2}<\cdots<a_{n-k}\leq n-1}a_{1}a_{2}\cdots a_{n-k} \] とおくと、
\begin{align*} \left(-1\right)^{n+k}A\left(n,k\right) & =\sum_{1\leq a_{1}<a_{2}<\cdots<a_{n-k}\leq n-1}a_{1}a_{2}\cdots a_{n-k}\\ & =\sum_{1\leq a_{1}<a_{2}<\cdots<a_{n-1-k}\leq a_{n-k}-1\land n-k\leq a_{n-k}\leq n-1}a_{1}a_{2}\cdots a_{n-k}\\ & =\sum_{1\leq a_{1}<a_{2}<\cdots<a_{n-1-k}\leq j-1\land n-k\leq j\leq n-1}a_{1}a_{2}\cdots a_{n-k-1}j\\ & =\sum_{j=n-k}^{n-1}j\sum_{1\leq a_{1}<a_{2}<\cdots<a_{n-1-k}\leq a_{n-k}-1}a_{1}a_{2}\cdots a_{n-k-1}\\ & =\sum_{j=n-k}^{n-1}j\sum_{1\leq a_{1}<a_{2}<\cdots<a_{n-1-k}\leq j-1}a_{1}a_{2}\cdots a_{j-\left(j-n+k+1\right)}\\ & =\sum_{j=n-k}^{n-1}j\left(-1\right)^{j-j+n-k-1}A\left(j,j-n+k+1\right)\\ & =\left(-1\right)^{n+k+1}\sum_{j=n-k}^{n-1}jA\left(j,j-n+k+1\right) \end{align*} となるので、
\begin{align*} A\left(n,k\right) & =-\sum_{j=n-k}^{n-1}jA\left(j,j-n+k+1\right)\\ & =-\sum_{j=0}^{k-1}\left(j+n-k\right)A\left(j+n-k,j+1\right)\\ & =-\sum_{j=1}^{k}\left(j+n-k-1\right)A\left(j+n-k-1,j\right) \end{align*} \(n\rightarrow n+k+1\)とすると、
\begin{align*} A\left(n+k+1,k\right) & =-\sum_{j=1}^{k}\left(n+j\right)A\left(n+j,j\right)\\ & =-\sum_{j=0}^{k}\left(n+j\right)A\left(n+j,j\right)+nA\left(n,0\right)\\ & =-\sum_{j=0}^{k}\left(n+j\right)A\left(n+j,j\right) \end{align*} となり第1種スターリング数の漸化式と同じになる。
しかし、\(\left(-1\right)^{0+k}A\left(0,k\right)=0\ne\delta_{0,k}=S_{1}\left(0,k\right),\left(-1\right)^{1+k}A\left(1,k\right)=0\ne\delta_{1,k}=S_{1}\left(1,k\right),\left(-1\right)^{2+k}A\left(2,k\right)=\delta_{2,k}-\delta_{1,k}=S\left(2,k\right)\)となるので、\(\left(-1\right)^{0+k}A\left(0,k\right),\left(-1\right)^{1+k}A\left(1,k\right)\)はスターリング数とは異なり、\(S_{1}\left(2,k\right)=\left(-1\right)^{2+k}A\left(2,k\right)\)となるので、\(2\leq n,0\leq k\leq n\)で\(S_{1}\left(n,k\right)=\left(-1\right)^{n+k}A\left(n,k\right)\)が成り立つ。
故に題意は成り立つ。

(2)

\(n,k\in\mathbb{N}_{0}\)として、
\[ A\left(n,k\right)=\sum_{a_{1}+a_{2}+\cdots+a_{k}=n-k}1^{a_{1}}2^{a_{2}}\cdots k^{a_{k}} \] とおくと、
\begin{align*} A\left(n,k\right) & =\sum_{a_{1}+a_{2}+\cdots+a_{k}=n-k}1^{a_{1}}2^{a_{2}}\cdots k^{a_{k}}\\ & =\sum_{a_{1}+a_{2}+\cdots+a_{k-1}=n-a_{k}-k}1^{a_{1}}2^{a_{2}}\cdots k^{a_{k}}\\ & =\sum_{a_{1}+a_{2}+\cdots+a_{k-1}=n-a_{k}-k}1^{a_{1}}2^{a_{2}}\cdots\left(k-1\right)^{a_{k-1}}k^{a_{k}}\\ & =\sum_{a_{1}+a_{2}+\cdots+a_{k-1}=n-j-k}1^{a_{1}}2^{a_{2}}\cdots\left(k-1\right)^{a_{k-1}}k^{a_{k}}\\ & =\sum_{a_{1}+a_{2}+\cdots+a_{k-1}=n-j-1-\left(k-1\right)}1^{a_{1}}2^{a_{2}}\cdots\left(k-1\right)^{a_{k-1}}k^{a_{k}}\\ & =\sum_{j=0}^{n-1}k^{a_{k}}\sum_{a_{1}+a_{2}+\cdots+a_{k-1}=n-j-1-\left(k-1\right)}1^{a_{1}}2^{a_{2}}\cdots\left(k-1\right)^{a_{k-1}}\\ & =\sum_{j=0}^{n-1}k^{j}A\left(n-j-1,k-1\right) \end{align*} となり、第2種スターリング数の漸化式と同じになる。
また初期条件も\(A\left(0,k\right)=\delta_{0,k}=S_{2}\left(0,k\right)\)となるので、\(A\left(n,k\right)=S_{2}\left(n,k\right)\)となる。
故に題意は成り立つ。