The set partitions of a finite set form a finite set

If $n=0$, then $[n]=\mathrm{\varnothing }$, whose only partition is the empty family. So the statement holds in the empty-set boundary case.

Assume $n\ge 1$. For a partition $\mathcal{P}$ of $[n]$, define $f_{\mathcal{P}}:[n]\to [n]$ by sending each $i\in [n]$ to the least element of the block of $\mathcal{P}$ containing $i$. If $i$ and $j$ lie in the same block, then they have the same least block element. Conversely, if $f_{\mathcal{P}}(i)=f_{\mathcal{P}}(j)$, then both $i$ and $j$ lie in the block containing that common least element, so they lie in the same block. Thus $\mathcal{P}$ is recovered from the fibres of $f_{\mathcal{P}}$.

Therefore $\mathcal{P}\mapsto f_{\mathcal{P}}$ is injective from the set of partitions of $[n]$ into the function set $[n{]}^{[n]}$. By The set $A^B$ of functions $B\to A$ between finite sets is finite, with $\mid A^B\mid =\mid A{\mid }^{\mid B\mid }$, the latter is finite. Hence the set of all partitions of $[n]$ is finite, and every subfamily of it, including the partitions with exactly $k$ blocks, is finite as well.