Easy one

$\dfrac{4}{(\sqrt{5}+1).(\sqrt[4]{5}+1).(\sqrt[8]{5}+1).(\sqrt[16]{5}+1)} = \dfrac{4}{(\sqrt{5}+1).(\sqrt[4]{5}+1).(\sqrt[8]{5}+1).(\sqrt[16]{5}+1)} \cdot \dfrac{\sqrt[16]{5}}{\sqrt[16]{5}} \\ =\dfrac{4(\sqrt[16]{5}-1)}{(\sqrt{5}+1).(\sqrt[4]{5}+1).(\sqrt[8]{5}+1).(\sqrt[8]{5}-1)} \quad \quad \quad [\because (\color{#20A900}{a}+\color{#3D99F6}{b})(\color{#20A900}{a}-\color{#3D99F6}{b})=\color{#20A900}{a}^2-\color{#3D99F6}{b}^2] \\ = \dfrac{4(\sqrt[16]{5}-1)}{(\sqrt{5}+1).(\sqrt[4]{5}+1).(\sqrt[4]{5}-1)} = \dfrac{4(\sqrt[16]{5}-1)}{(\sqrt{5}+1).(\sqrt{5}+1)} =\dfrac{4(\sqrt[16]{5}-1)}{4} =\sqrt[16]{5}-1 \\ \implies x=\sqrt[16]{5}-1 \therefore x+1=\sqrt[16]{5} \\ \therefore (x+1)^{48}=5^{\frac{48}{16}} =5^3 \\ =\boxed{125}$

Let $k=\sqrt[16]{5}:$

so $x=\frac{4}{(k^8+1)(k^4+1)(k^2+1)(k+1)}$

$x=\frac{4}{k^{15}+k^{14}+k^{13}+k^{12}+k^{11}+k^{10}+k^9+k^8+k^7+k^6+k^5+k^4+k^3+k^2+k+1}$

$x=\frac{4(k-1)}{k^{16}-1}$ replacing $k=\sqrt[16]{5}$ so: $x=\sqrt[16]{5}-1$ so: $(x+1)^{48}=k^{48}=125.$