Let $!$ be the D6: Natural number factorial function.
Let $n \in \mathbb{N}$ be a D996: Natural number.
Let $n \in \mathbb{N}$ be a D996: Natural number.
Then
(1) | \begin{equation} n ! \leq n^n \end{equation} |
(2) | \begin{equation} n! = n^n \quad \iff \quad n \in \{ 0, 1 \} \end{equation} |