Revised after the suggestion of @Metamorphy We wish to sum: $$S_{n,m}=\sum_{k=1}^{n} (-1)^k~ k^{m} ~ {n \choose k}~\text{ for}, m<n,=,>n.$$ Let us consider an interesting function $$f(x)=(1-e^{x})^n = 1~-~n e^{x}+\frac{n(n-1)}{2!} e^{2x}-\frac {n(n-1)(n-2)}{3!} e^{3x}+...+(-1)^n e^{nx}~~~~ (2)$$ Next the coefficient of $x^{m}$ in the RHS of (2) is $$-\frac{n}{1!} \frac{1^{m}}{m!}+\frac{n(n-1)}{2!} \frac{2^{m}}{m!}-\frac{n(n-1)(n-2)}{3!} \frac{3^{m}}{m!}+....=-\frac{S_{n,m}}{m!} ...(3).$$ Also, the coefficient of $x^{m}$ in $f(x)$ can be obtained from $$f(x)= (-1)^n (e^x-1)^n = (-1)^n \left (x^n+\frac{n}{2}x^{n+1}+\frac{1}{24}(n+3n^2) x^{n+2}...\right)...(4)$$ (McLaurin expansions), where there are terms having higher powers of $x$, namely $x^n$ on wards.
Three cases arise $$S_{n,m<n}=0, m<n...(5).$$ $$S_{n,m=n}=(-1)^n ~ n!.....(6).$$ $$S_{n,n+1}=(-1)^n \frac{1}{2}(n+1)!....(7).$$ $$S_{n,n+2}=(-1)^n \frac{1}{24}~ n~(1+3n)~(n+2)!....(8)$$
But Mathematica gives this answer as $$S_{n,n+1}(-1)^n \frac{1}{2} n(n+1)! ~.$$ Next, from Eqs. (3) and (4), we get $$S_{n,n+2}=(-1)^n \frac{1}{24}~(3n+1)~(n+3) ~(n+2)!....(8).$$
The LHS of $(3)$ is not $-\dfrac{S_{n,m}}{m!}$ but rather $\displaystyle\frac{1}{m!}\sum_{k=\color{red}{0}}^{n}(-1)^k\color{red}{(}k\color{red}{{}+1)}^m\binom{n}{k}$.
To get to $S_{n,m}$, consider just $f(x)=(1-e^x)^n$ instead.