We can start from a standard Stirling number identity:
$$x^r = \sum_{j=0}^{r} (-1)^{r-j} \left\{\!{r \atop j}\!\right\} x^{(j)} = \sum_{j=0}^{r} (-1)^{r-j} j!\left\{\!{r \atop j}\!\right\} \binom{x+j-1}{j}.$$
We can make this look more like the identity in the question by shifting the sum so that $j=k-1$, and substituting $x = N-n$. (It also looks like we want $r=K-1$, but I am waiting on that step for a reason.) But when we make this substitution, the result is $$x^r = (N-n)^r = \sum_{k=1}^{r+1} (-1)^{r-k+1} k!\left\{\!{r \atop k-1}\!\right\} \binom{N-n+k-2}{k-1}.$$ The binomial coefficients are the right ones! But on the left, we have $(N-n)^r$ instead of $n^{K-1}$.
To fix this, let's expand $n^{K-1}$ in terms of powers of $x$: $$n^{K-1} = (N-x)^{K-1} = \sum_{r=0}^{K-1} (-1)^r\binom{K-1}{r} N^{K-1-r} x^r.$$ This lets us substitute the sum for $x^r$ to get $$n^{K-1} = \sum_{r=0}^{K-1} (-1)^r\binom{K-1}{r} N^{K-1-r} \sum_{k=1}^{r+1} (-1)^{r-k+1} k!\left\{\!{r \atop k-1}\!\right\} \binom{N-n+k-2}{k-1}.$$ Now, swap the order of summation! Instead of $r=0$ to $K-1$ and $k=1$ to $r+1$, we'll take $k=1$ to $K$ and $r=k-1$ to $K-1$. Cleaning things up a little, we get $$n^{K-1} = \sum_{k=1}^K \color{red}{\left((-1)^{k+1} k! \sum_{r=k-1}^{K-1} \binom{K-1}{r} N^{K-1-r} \left\{\!{r \atop k-1}\!\right\}\right)} \binom{N-n+k-2}{k-1}$$ which gives us the coefficient $$c_k = (-1)^{k+1} k! \sum_{r=k-1}^{K-1} \binom{K-1}{r} N^{K-1-r} \left\{\!{r \atop k-1}\!\right\}.$$
It's possible this simplifies further: the expression for $c_k$ looks a lot like the identity $$\left\{\!{K \atop k}\!\right\} = \sum_{r=k-1}^{K-1} \binom{K-1}{r} \left\{\!{r \atop k-1}\!\right\}$$ except that there's a power of $N$ in the sum as well.
