12.14 Given the complex sequence $\left\{a_{n}\right\}$ with the general term
$$
a_{n}=(1+i)\left(1+\frac{i}{\sqrt{2}}\right)\left(1+\frac{i}{\sqrt{3}}\right) \cdots\left(1+\frac{i}{\sqrt{n}}\right) .
$$

Then $\left|a_{n}-a_{n+1}\right|$ equals
(A) $\frac{\sqrt{2}}{2}$.
(B) $\sqrt{2}$.
(C) 1 .
(D) 2 .
(High School Mathematics Competition, Suzhou, Jiangsu Province, China, 1993)
[Solution] First, prove that $\left|a_{n}\right|=\sqrt{n+1}$ using mathematical induction.
(1) When $n=1$, $\left|a_{1}\right|=|1+i|=\sqrt{1+1}$, the conclusion holds.
(2) Assume that when $n=k$, $\left|a_{k}\right|=\sqrt{k+1}$ holds. Then, when $n=k+1$,
$$
\left|a_{k+1}\right|=\left|a_{k} \cdot\left(1+\frac{i}{\sqrt{k+1}}\right)\right|=\left|a_{k}\right| \sqrt{1+\frac{1}{k+1}}=\sqrt{k+2} \text {. }
$$

Thus, for all $n \in \mathbb{N}, \left|a_{n}\right|=\sqrt{n+1}$.
$$
\begin{aligned}
\therefore \quad & \left|a_{n}-a_{n+1}\right| \\
= & \left|a_{n}-a_{n}\left(1+\frac{i}{\sqrt{n+1}}\right)\right| \\
= & \left|a_{n} - a_{n} \cdot \frac{i}{\sqrt{n+1}} \right| \\
= & \left|a_{n}\right| \cdot\left|\frac{i}{\sqrt{n+1}}\right| \\
= & \sqrt{n+1} \cdot \frac{1}{\sqrt{n+1}} \\
= & 1 .
\end{aligned}
$$

Therefore, the answer is $(C)$.