Ooh, kinda cool problem I think this inequality relies on your interpretation of each side. Consider labelling each of the N squares 1 through N. The left hand side gives the number of ways to permute the labels on each of the rows, and then with these new labels permute the labels on each of the columns. Each way you could do this clearly gives some unique labelling of the squares. The RHS gives the number of ways to permute the labels on the squares in general. Therefore, since the permuting process described for the LHS is clearly a subset of all ways to permute the N labels, the inequality holds.

Proposed by Fedor Petrov. Really,nice problem.

In a way this question also gives a really nice identity that if $\sum a_i=\sum b_i=N$ and $p$ is an arbitary prime then $s_p(N)+N\le \sum s_p(a_i) + \sum s_p(b_i)$ where $s_p$ is sum of digits of that number in base $p$.This interpretation of the question follows from Legendre's Formula

Isn't it just an induction? Use this: $a_1...a_n \leq \binom{a_1+...+a_n}{n}$