Is there another way to write the product $ 2^2\times 3^2\times 4^2\times 5^2\times 6^2\times \cdots =(n^2)!\;$?

Question

I have this product

$$ 2^2\times 3^2\times 4^2\times 5^2\times 6^2\times \cdots =(n^2)!$$

Can we write this in another way?

Thank you

Answer 1

That's not $(n^2)!$, it is $(n!)^2$.

(For example, if $n=3$ then $(n^2)!$ is $(3^2)!=9!=362880$ whereas $(n!)^2=(3!)^2=6^2=36$.)

Answer 2

Well in product notation it is $$\prod_{k=1}^n k^2$$, but realistically $(n!) ^2$ is much better notation.