You can also use the following finite calculus formula for alternating binomial transforms. Let $B(f(k),n)$ denote the alternating binomial transform; i.e., $B(f(k),n) = \sum_{k=0}^n (-1)^k \binom{n}{k} f(k).$ Then (the formula) $B(f(k),n) = -B(\Delta f(k),n-1) + f(0)[n=0].$ (Here, $\Delta f(k)$ is the finite difference $f(k+1) - f(k)$, and the expression $[n=0]$ evaluates to $1$ if $n = 0$ and $0$ otherwise. Also, note that the first argument to $B$ is a function, while the second is a number.)
Starting with $f(k) = 1$, we have $\Delta f(k) = 0$. Since $B(0,n-1)$ is clearly $0$, $B(1,n) = [n=0]$. The latter is just the known formula $\sum_{k=0}^n (-1)^k \binom{n}{k} = [n=0].$
Then, continuing to take antidifferences, and using the notation $k^{\underline{m}}$ for the falling factorial $k(k-1)\cdots (k-m+1)$ as well as the power rule for finite differences $\Delta k^{\underline{m}} = m k^{\underline{m-1}}$, we have
$\sum_{k=0}^n (-1)^k \binom{n}{k}k = -B(1,n-1) = - [n-1=0] = -[n=1],$ $\sum_{k=0}^n (-1)^k \binom{n}{k} k^{\underline{2}} = -2B(k,n-1) = 2[n-1 = 1] = 2[n=2],$ $\sum_{k=0}^n (-1)^k \binom{n}{k} k^{\underline{3}} = -3B(k^{\underline{2}},n-1) = -6[n-1 = 2] = -6[n=3],$ and so forth, until we get to $\sum_{k=0}^n (-1)^k \binom{n}{k} k^{\underline{m}} = -mB(k^{\underline{m-1}},n-1) = (-1)^m m![n=m].$
Since $k(k-1)\cdots (k-m+1) = \frac{k!}{(k-m)!}$, dividing both sides of this last identity by $(-1)^m m!$ proves the OP's identity.
(The finite calculus formula for alternating binomial transforms and this argument are excerpted from my paper "Combinatorial sums and finite differences," Discrete Mathematics 307 (24): 3130-3146, 2007.)