7. What is the second digit (from the left) of the number $\left(10^{16}+1\right)\left(10^{8}+1\right)\left(10^{4}+1\right)\left(10^{2}+1\right)(10+1)$?
(A) 0
(B) 1
(C) 2
(D) 3
(E) 4.
7. The answer is $\mathbf{( B )}$. All the digits of the number can be calculated directly. We have $10^{16}+1=$ $\frac{10^{32}-1}{10^{16}-1}$ (it is the notable product $(a-1)(a+1)=a^{2}-1$), and similar factorizations for the other terms. Therefore,

$$
\begin{aligned}
& \left(10^{16}+1\right)\left(10^{8}+1\right)\left(10^{4}+1\right)\left(10^{2}+1\right)(10+1) \\
= & \frac{10^{32}-1}{10^{16}-1} \frac{10^{16}-1}{10^{8}-1} \frac{10^{8}-1}{10^{4}-1} \frac{10^{4}-1}{10^{2}-1} \frac{10^{2}-1}{10-1} \\
= & \frac{10^{32}-1}{9}=\overbrace{\frac{399 \ldots 9}{9}}^{9}=\overbrace{111 \ldots 1}^{32 \text { digits }} .
\end{aligned}
$$