$
\large\begin{array}{l}
\frac{{(n!)^2 }}{{n^2 }} = \\
\frac{{\left( {1 \cdot 2 \cdot 3 \cdot ... \cdot n} \right)^2 }}{{n^2 }} = \\
\frac{{1^2 \cdot 2^2 \cdot 3^2 \cdot ... \cdot n^2 }}{{n^2 }} = \\
\end{array}
$
$
\begin{array}{l}
1^2 \cdot 2^2 \cdot 3^2 \cdot ... \cdot \left( {n - 1} \right)^2 = \\
(1 \cdot 2 \cdot 3 \cdot ... \cdot \left( {n - 1} \right))^2 = \\
(n - 1)!^2 \\
\end{array}
$
Viel mee, denk ik...
WvR
5-10-2013