ขอไอเดียพิสูจน์หน่อยครับ
 

1.Let $n\ge2$. Then
$$\sum_{k = 1}^{n} (-1)^k k^j\binom{n}{k} =0, \;\;\; for\;\; j=1,2,...,n-1.$$

Define an operator $\Delta$ by
$$\Delta p(x):=p(x+1)-p(x)$$
for all polynomials $p(x)\in\mathbb{C}[x]$. Prove that, if $p(x)\in\mathbb{C}[x]$ has degree $n$, then $$\Delta^{n+1}p(x)=0\tag{*}\,.$$ Now, observe that
$$\sum_{k=0}^n\,(-1)^{n-k}\,\binom{n}{k}\,(x+k)^j=\Delta^np_j(x)\,,$$
where $p_j(x):=x^j$ for $j=0,1,2,3,\ldots$. By the observation (*), we conclude that $$\sum_{k=0}^n\,(-1)^{n-k}\,\binom{n}{k}\,(x+k)^j=0$$ for $j=0,1,2,\ldots,n-1$. Plugging in $x:=0$, we obtain $$(-1)^n\,\sum_{k=0}^n\,(-1)^k\,k^j\,\binom{n}{k}=\sum_{k=0}^n\,(-1)^{n-k}\,\binom{n}{k}\,k^j=0$$ for every $j=0,1,2,\ldots,n-1$ (we use the convention $0^0=1$ here). Note, however, that
$$\sum_{k=0}^n\,(-1)^{n-k}\,\binom{n}{k}\,k^n=n!\,.$$

More generally, let $\displaystyle{p \brace q}$ be the Stirling number of the second kind with parameters $p,q\in\mathbb{Z}_{\geq 0}$. We can write
$$x^p=\sum_{r=0}^n\,{p \brace r}\,r!\,\binom{x}{r}$$
for all $p\in\mathbb{Z}_{\geq 0}$. Therefore,
$$\sum_{k=0}^n\,(-1)^{n-k}\,\binom{n}{k}\,k^j=\sum_{k=0}^n\,(-1)^{n-k}\,\binom{n}{k}\,\sum_{r=0}^j\,{j \brace r}\,r!\,\binom{k}{r}\,.$$
That is,
$$\sum_{k=0}^n\,(-1)^{n-k}\,\binom{n}{k}\,k^j=\sum_{r=0}^j\,{j \brace r}\,r!\,\sum_{k=0}^n\,(-1)^{n-k}\,\binom{n}{k}\,\binom{k}{r}\,.$$
When $r<n$, then
$$\sum_{k=0}^n\,(-1)^{n-k}\,\binom{n}{k}\,\binom{k}{r}=\sum_{k=r}^n\,(-1)^{n-k}\,\binom{n}{r}\,\binom{n-r}{k-r}=\binom{n}{r}\,(1-1)^{n-r}=0\,.$$
When $r=n$, then
$$\sum_{k=0}^n\,(-1)^{n-k}\,\binom{n}{k}\,\binom{k}{r}=\binom{n}{n}\,\binom{n}{n}=1\,.$$
When $r>n$, then
$$\sum_{k=0}^n\,(-1)^{n-k}\,\binom{n}{k}\,\binom{k}{r}=\sum_{k=0}^n\,(-1)^{n-k}\,\binom{n}{k}\cdot 0=0\,.$$
Therefore,
$$\sum_{k=0}^n\,(-1)^{n-k}\,\binom{n}{k}\,k^j={j\brace n}\,n!\,,$$
where we note that $\displaystyle{j\brace n}$ is conventionally defined to be $0$ if $j<n$.