Proof
(related to Proposition: Comparison between the Stirling numbers of the First and Second Kind)
 Let $n,r\ge 0$ be integers.
 The number of partitions of $n$ objects into $r$ nonempty subsets corresponds to the Stirling numbers of the second kind $\left\{\begin{array}{c}n\\r\end{array}\right\}.$
 Every partition of $n$ objects into $r$ nonempty subsets leads to at least one partition of $n$ objects into $r$ cycles.
 Since the number of partitions of $n$ objects into $r$ cycles corresponds to the Stirling numbers of the first kind $\left[\begin{array}{c}n\\r\end{array}\right],$ it follows $\left[\begin{array}{c}n\\r\end{array}\right]\ge \left\{\begin{array}{c}n\\r\end{array}\right\}$ for all $n,r\ge 0.$
 The number of arrangements of $n$ objects into $n$ nonempty subsets and $n$ cycles equal each other and is $1$
 Therefore, $\left[\begin{array}{c}n\\n\end{array}\right]=\left\{\begin{array}{c}n\\n\end{array}\right\}=1.$
 In any arrangement of $n$ objects into $n1$ nonempty subsets there is exactly one subset with $2$ elements and all remaining subsets are singletons.
 Moreover, all these subsets with either $1$ or $2$ elements correspond to cycles and there are $\binom n2=\frac{n(n1)}{2}$ such arrangements.
 Therefore, $\left[\begin{array}{c}n1\\n\end{array}\right]=\left\{\begin{array}{c}n\\n1\end{array}\right\}=\left(\begin{array}{c}n\\2\end{array}\right)=\frac{n(n1)}{2}.$
∎
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References
Bibliography
 Graham L. Ronald, Knuth E. Donald, Patashnik Oren: "Concrete Mathematics", AddisonWesley, 1994, 2nd Edition