Divizibilitate

[Mateescu Constantin]

Demonstrati ca \( \(2^{2x}+2^{x+y}+2^{2y}\)!\ \vdots\ \(2^x!\)^{2^x+2^{y-1}}\ \cdot\ \(2^y!\)^{2^y+2^{x-1}}\ ,\ \forall\ x,y\in\mathbb{N}^{\ast} \) .