Maximizing a product of factorials I would like to maximize $n_1! n_2! \cdots n_k!$ under the constraint $n_1 + n_2 + \cdots + n_k = N$ and $n_i &gt; 0$ for all $i$. Intuitively, I think the maximum occurs when all $n_i$ are $1$ except for one of them: $$(1!)( 1!) \cdots (1!)( (N-(k-1))!)$$ but I am unsure how to show this because I can't take the derivative of this kind of function. Could someone point me in the right direction? Thanks!



Edit: perhaps induction on $k$ will work?

If $k=2$, it is a little easier to see that the maximum of $n_1!n_2!=n_1! (N-n_1)!$ is $N!$ (how to rigorously show this still escapes me). Then it follows by induction, I believe.
Increasing the largest $n_i$ by 1 and simultaneously decreasing the smallest $n_j$ by 1 respects the condition $n_1 + n_2 + \cdots + n_k = N$ and increases the product.  Therefore you can make all of the factors equal to 1 except the biggest one.
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$\ds{{\cal F} \equiv \sum_{\ell = 1}^{k}\ln\pars{\Gamma\pars{n_{\ell} + 1}} - \mu\pars{\sum_{\ell = 1}^{k}n_{\ell} - N}}$
$$
0 = \partiald{{\cal F}}{\tilde{n}_{\ell}} = \Psi\pars{\tilde{n}_{\ell} + 1} - \mu\,,
\quad\forall\ \ell\quad\imp\quad \tilde{n}_{\ell} = {N \over k}\,,\quad\forall\ \ell
$$