$1+\frac{1}{x}+\frac{2(x+1)}{xy}+\frac{3(x+1)(y+2)}{xyz}+\frac{4(x+1)(y+2)(z+3)}{xyzw}=\frac{(x+1)(y+2)}{xy}+\frac{3(x+1)(y+2)}{xyz}+\frac{4(x+1)(y+2)(z+3)}{xyzw}=\frac{(x+1)(y+2)}{xy} \cdot \left[1+\frac{3}{z}+\frac{4(z+3)}{zw} \right]=\frac{(x+1)(y+2)}{xy} \cdot \frac{(z+3)(w+4)}{zw}=\frac{(x+1)(y+2)(z+3)(w+4)}{xyzw}.$ So we get the equation $\frac{(x+1)(y+2)(z+3)(w+4)}{xyzw}=0$ which implies $x=-1$ or $y=-2$ or $z=-3$ or $w=-4$.