Solution:
The answer is $\mathbf{( B )}$. One can directly compute all the digits of the number. We have $10^{16}+1=\frac{10^{32}-1}{10^{16}-1}$ (this is the well-known identity $(a-1)(a+1)=a^{2}-1$ ), and analogous decompositions for the other terms. Hence
$$\begin{aligned} & \left(10^{16}+1\right)\left(10^{8}+1\right)\left(10^{4}+1\right)\left(10^{2}+1\right)(10+1) \\ = & \frac{10^{32}-1}{10^{16}-1} \frac{10^{16}-1}{10^{8}-1} \frac{10^{8}-1}{10^{4}-1} \frac{10^{4}-1}{10^{2}-1} \frac{10^{2}-1}{10-1} \\ = & \frac{10^{32}-1}{9}=\overbrace{\frac{399\ldots 9}{9}}^{9}=\overbrace{111\ldots 1}^{32 \text{ digits}} . \end{aligned}$$