        \problemdata       % Takes 5 *mandatory* arguments
        {mult-nfracs}             % latex-friendly label for the prob.
        {induction}  % The topic(s) of the problem content.
        {}               % Source (if known)
        {F01 PS2-1} % Usage (list ones you are aware of).
        {}        % Last revision info (author, date).

\begin{problem}
 Use induction to prove that the following equation holds for all
$n \geq 2$.

\begin{equation*}
\left(1 - \frac{1}{2}\right)
\left(1 - \frac{1}{3}\right)
\cdots
\left(1 - \frac{1}{n}\right)  =  \frac{1}{n}
\end{equation*}

\solution{\begin{proof}
The proof is by induction on $n$.  Let $P(n)$ be the proposition that
the above equation holds.  In the base case, $P(2)$ is true because
$(1 - \frac{1}{2}) = \frac{1}{2}$.  In the inductive step, for $n \geq
2$ assume $P(n)$ to prove $P(n+1)$ as follows.

\begin{eqnarray*}
\left(1 - \frac{1}{2}\right) \left(1 - \frac{1}{3}\right)
\cdots
\left(1 - \frac{1}{n}\right) \left(1 - \frac{1}{n+1}\right)
        & = &   \frac{1}{n} \left(1 - \frac{1}{n+1}\right) \\
        & = &   \frac{1}{n} \cdot \frac{n}{n+1} \\
        & = &   \frac{1}{n+1}
\end{eqnarray*}

The first step uses the induction hypothesis, and the second and third
steps are simplifications.  This shows that $P(n)$ implies $P(n+1)$,
and the claim is proved by induction.
\end{proof}
}

\end{problem}