The Stirling numbers of the first and the second kind obey the following inequality:
$$\left[\begin{array}{c}n\\r\end{array}\right]\ge \left\{\begin{array}{c}n\\r\end{array}\right\},\quad n,r\ge 0.$$ Equality holds for $r=n$ and for $r=n-1,$ in particular $$\left[\begin{array}{c}n\\n\end{array}\right]=\left\{\begin{array}{c}n\\n\end{array}\right\}=1,\quad \left[\begin{array}{c}n\\n-1\end{array}\right]=\left\{\begin{array}{c}n\\n-1\end{array}\right\}=\binom n{2}=\frac{n(n-1)}{2}.$$
Proofs: 1
