# Explanation for the $\frac{(n!)^{2n}}{n^{n^2}}$ lower bound on the number of Latin squares?

Wikipedia says that there are at least $$\frac{(n!)^{2n}}{n^{n^2}}$$ Latin squares of size $$n$$. But the citation is a paywalled textbook. How does one prove this bound?

Here is a sketch of their proof. Build your Latin square one row at a time. At the $$k^{th}$$ stage, you have a $$k\times n$$ Latin rectangle, $$R_k$$. Let us count the number of ways to extend $$R_k$$ to $$R_{k+1}$$. This can be computed as the permanent of a certain matrix. Namely, for each $$k\in \{0,1,\dots,n-1\}$$, define the $$n\times n$$ matrix $$B^k$$ by saying $$B^k_{i,j}=1$$ if the $$j^{th}$$ column of $$R_k$$ does not contain the entry $$i$$. Then the number of ways to extend $$R_k$$ is $$\def\per{\text{per }}\per B_k$$. Therefore, $$\text{# n\times n Latin squares} \ge \prod_{k=0}^{n-1}. \per B_k\tag1$$ Finally, you can show that $$\frac{1}{n-k} B_k$$ is a doubly stochastic matrix, which by Van der Waerden's conjecture implies $$\per \!\!\left(\frac{1}{n-k}B_k\right)\ge n!/n^n$$, or $$\per B_k \ge (n-k)^n n!/n^n.\tag 2$$ Combining $$(1)$$ and $$(2)$$ gives you what you want.