Prove $n! = o(n^n)$ without limits
You do not need anything as fancy as Stirling. It is easy to see that $n!\cdot n &lt; n^n$ for all $n&gt;2$ (just compare the first two terms of $n!=1\cdot 2\cdot\ldots$ witht those of $n^n=n\cdot n\cdot\ldots$).