Factorials, squares and Bertrand's postulate With Bertrand's postulate at hand, it is easy to see that $n!$ is never a square for $n\ge 2$ (because there is a prime between $n/2$ and $n$).
But are there more elementary proofs of that fact? 
 A: Not an answer just some speculative thoughts.
Let's define $T_n$ as the n'th tetrahedral number, where
$$T_n=\frac{1}{6}n(n+1)(n+2)$$
The even tetrahedral numbers $T_{2n}$ are also given by
$$T_{2n}=2^2+4^2+6^2+...+(2n)^2$$
The odd tetrahedral numbers $T_{2n-1}$ are also given by
$$T_{2n-1}=1^2+3^2+5^2+...+(2n-1)^2$$
We can define the factorial in terms of a tetrahedral number thus
$$n!=6(n-3)! \;T_{n-2}$$ 
We can then iterate this process as long as the residual factorial left over is 2! or greater. The factorial is then calculated from the product of two or more tetrahedral numbers, alternating between odd and even tetrahedral numbers.
$$n!=6^2(n-6)! \;T_{n-5}\;T_{n-2}$$ 
$$n!=6^3(n-9)! \;T_{n-8}\;T_{n-5}\;T_{n-2}$$ 
According to Dickson (History of the Theory of Numbers (Vol II. p.25)) A.J.J. Meyl [REF1] has proved that only three tetrahedral numbers are perfect squares; that is $T_1=1^2$, $T_2=2^2$ and $T_{48}=140^2$. Assuming it is correct it is an elementary proof.
The question is whether it is possible to build on this proof and the properties of tetrahedral numbers, to determine when the product of odd and even tetrahedral numbers are squares or not squares.  
[REF1] Meyl, A.J.J. "Solution de Question 1194." Nouv. Ann. Math. 17, 464-467, 1878. http://www.numdam.org/article/NAM_1878_2_17__464_1.pdf
