Is it very big or very small?

$\begin{aligned} &\large \left ( \dfrac{4}{(\sqrt{5}+1)(\sqrt[4]{5}+1)(\sqrt[8]{5}+1)(\sqrt[16]{5}+1)} + 1\right ) ^{48}\\ &=\left ( \dfrac{4(\sqrt[16]{5}-1)}{(\sqrt{5}+1)(\sqrt[4]{5}+1)(\sqrt[8]{5}+1)(\sqrt[16]{5}+1)(\sqrt[16]{5}-1)} + 1\right ) ^{48}\\ &=\left ( \dfrac{4(\sqrt[16]{5}-1)}{(\sqrt{5}+1)(\sqrt[4]{5}+1)(\sqrt[8]{5}+1)(\sqrt[8]{5}-1)} + 1\right ) ^{48}\\ &=\left ( \dfrac{4(\sqrt[16]{5}-1)}{(\sqrt{5}+1)(\sqrt[4]{5}+1)(\sqrt[4]{5}-1)} + 1\right ) ^{48}\\ &=\left ( \dfrac{4(\sqrt[16]{5}-1)}{(\sqrt{5}+1)(\sqrt{5}-1)} + 1\right ) ^{48}\\ &=\left ( \dfrac{4(\sqrt[16]{5}-1)}{5-1} + 1\right ) ^{48}\\ &=(\sqrt[16]{5})^{48}\\ &=5^3\\ &=125 \end{aligned}$

Thus, the answer is 125.

I started by working on the denominator. Let $G = \sqrt[16]{5}$ , then the denominator is $D = (G^8+1)(G^4+1)(G^2+1)(G+1).$ Expanding the brackets, we get 16 terms, which turn out to be exactly $D = G^0 + G^1 + \cdots + G^{15}.$ Using the fact that $\sum_{i=0}^{n-1} x^i = (1 - x^n)/(1-x)$ , $D = \frac{G^{16} - 1}{G - 1} = \frac{5-1}{\sqrt[16]{5} - 1}.$ The rest is easy: $\left(\frac 4 D + 1\right)^{48} = (\sqrt[16]{5}-1+1)^{48} = (5^{1/16})^{48} = 5^3 = \boxed{125}.$

That was canny. ❵:-D