An algebra problem by Saurav Pal

We note that in general,

$\sqrt[2^n]{5} + 1)(\sqrt[2^n]{5} - 1) = (\sqrt[2^n]{5})^2 - 1^2 = \sqrt[2^{n-1}]{5} - 1$ .

It now becomes apparent that if we multiply the numerator and denominator of $\dfrac{4}{ (\sqrt{5}+1) (\sqrt[4]{5}+1) (\sqrt[8]{5}+1) (\sqrt[16]{5}+1) }$ by $(\sqrt[16]{5} - 1)$ , the denominator will telescope to $\sqrt[1]{5} - 1 = 4$ , so

$x = \dfrac{4(\sqrt[16]{5} - 1)}{4} = \sqrt[16]{5} - 1$

It follows that $(x + 1)^{48} = (\sqrt[16]5)^{48} = 5^3 = \boxed{125}$