We prove, for all \(k \in \mathbb{N}\), we have
\[ \left(\frac{k}{e}\right)^k \;\leq\; k! \;\leq\; k^k. \]The upper bound is trivial. For the lower bound we expand \(e^k\):
\[ e^k = \sum_{m=0}^\infty \frac{k^m}{m!} \;\;\geq\;\; \frac{k^k}{k!}. \]This yields the lower bound.