You want to count the number of set partitions of a set of $n$ elments, into $n/k$ parts each of size $k$.
(It is assumed that $k$ divides $n$.)
Method 1.
We can generate such a partition by writing down the $n$ elements in a sequence, and then declaring that the first $k$ elements are the first part, the next $k$ elements are the second part, and so on. There are $n!$ ways of writing $n$ elements in a sequence, but each partition is generated multiple times: for each of the $n/k$ parts, there are $k!$ orderings of the $k$ elements in that part that would lead to the same partition, as you don't care about the order within each part. Further, there are $(n/k)!$ orderings of the parts themselves, for the same partition.
The number of partitions is therefore:
$$\frac{n!}{(k!)^{n/k} (n/k)!}$$
Method 2.
You can choose the elements of the first part in $\binom{n}{k}$ ways, then choose the elements of the second part as $k$ out of the remaining $n-k$ in $\binom{n-k}{k}$ ways, and so on. But as different orderings of the $(n/k)$ parts don't change the partition, the number of partitions is
$$\frac{\binom{n\vphantom{k}}{k}\binom{n-k}{k}\cdots\binom{k}{k}}{(n/k)!} = \frac{n!}{(k!)^{n/k}(n/k)!}$$
as before.
You can verify that this accords with all your cases. For instance, for $n=15$ and $k=5$, you get $\frac{15!}{5!^3 3!} = 126126$. These numbers are tabulated in OEIS A060540, and no simpler formula is listed.
To come to a result you need a slightly different setup. For a fixed cardinality of a subset it is important to know how many subsets in the arrangement have this cardinality. These subsets cannot be distinghuished on base of cardinality so we must be aware of the "danger" of multiple counting.
Let's say that $N=r_1.m_1+\cdots+r_k.m_k$ where $m_i$ corresponds with the cardinality of the subset and $r_i$ stands for the number of times that a subset of this cardinality occurs in the partition.
E.g. $4=2.1+1.2$ corresponds with a splitup of $\{1,2,3,4\}$ in two singletons and one set of $2$ elements.
At first hand we find: $$\frac{N!}{\left(m_1!\right)^{r_1}\times\cdots\times\left(m_k!\right)^{r_k}}$$ possibilities, but this with multiple counting that must be repaired.
The endresult is:$$\frac1{r_1!\times\cdots\times r_k!}\frac{N!}{\left(m_1!\right)^{r_1}\times\cdots\times\left(m_k!\right)^{r_k}}$$
Let's return to example $4=2.1+1.2$. The first formula give $\frac{4!}{1!1!2!}=12$ possibilities. But here the possibilities $\{1\}\{2\}\{3,4\}$ and $\{2\}\{1\}\{3,4\}$ are thought of as "distinct". The second formula divides by $2$ to repair that. There are $6$ possibilities (each of them corresponds with a selection of $2$ elements out of $4$ that are "condemned" to be elements of the same subset and there are $\binom42=6$ possible selections).
Best Answer
One way to solve it: first put all $nk$ items in order: there are $(nk)!$ ways. Now chop them into blocks of $k$ and you have a partition. Each block can be reordered in $k!$ ways, then the blocks can be put in order in $n!$ ways. In all, we have $\frac{(nk)!}{(k!)^nn!}$ ways to make the partition.
Another way is to say there are ${nk \choose k}$ ways to get the first partition, ${(n-1)k \choose k}$ to get the second, and so on. Again you can choose the partitions in $n!$ orders. This gets you the same answer.