Let $P_N$ be the set of partitions of $N \in \mathbb{N}$. Denote an element of $P_N$ as
$$k=(n_1,n_2,...,n_N) \in P,$$
where $n_i \in \mathbb{N}$. I am wondering whether the partitions that maximise the function
$$f(k)=n_1!n_2! ... n_N!$$
are the ones with $n_i=N$ for some $i \in (1,...,N)$, and the rest zero. How does one go about proving this, if it is indeed true.
One approach to prove a result like this is to show that every solution that is not of this form can be improved. Suppose $n_i$ and $n_j$ are both positive, and replace $(n_i,n_j)$ with $(n_i+n_j,0)$. Then compare $f$ before and after this change.