I have the following exact sum for the expectation of an event

$$\sum_{m=0}^{nk} \sum_{j=1}^n (-1)^{j-1}\binom{n}{j} \binom{(n-j)k}{m} / \binom{nk}{m}$$

which is exactly correct but I want to give an upper bounding approximation that is easier to interpret. In particular, I want to see the expectations dependence on the parameter $k$. In simulation, it appears that as $k \rightarrow \infty$ we have that the sum tends towards $n \log n$. However, I am unsure how to show such a result analytically. My attempt has been to use some form of a stirling approximation on the binomial coefficients like $$\binom{n}{k} \leq \frac{n^k}{k!}$$ which can reduce the summation in the following way $$\sum_{m=0}^{nk} \sum_{j=1}^n (-1)^{j-1}\binom{n}{j} \frac{((n-j)k)^m}{m!} \cdot \frac{m!}{(nk)^m} = \sum_{m=0}^{nk} \sum_{j=1}^n (-1)^{j-1}\binom{n}{j} \left(1-\frac{j}{n}\right)^m$$ and from here maybe we can upper bound the $(1-j/n)^m$ to simplify further to a sum which depends only on $j$? Again comparing to simulation, it seems that when $k \ll \log n$ the sum will be something like $nk$ and for $k \gg \log n$ we will have the $n \log n$ (thus removing the dependence on $k$) but I can't seem to derive such a result.

Any help or advice would be *greatly* appreciated!