Why are strict inequalities stronger than non-strict inequalities? I'm working with induction proofs involving inequalities and I am encountering example proofs that wish to show things of the sort, $n!\le\ n^n$ for every positive integer. The proof given in the inductive step is, $(n+1)!$ $=$ $(n+1)\dot\ n!$ $\le$ $(n+1)n^n$$\lt$ $(n+1)(n+1)^n$ $=$ $(n+1)^{n+1}$ I'm not quite clear why the strict inequality was admissible in the proof, if I'm trying to show $n!\le\ n^n$, rather than $n!\lt\ n^n$. Thanks for any insight.

You can change $<$ by $\leq$ but not $\leq$ by $<$.
For example, we have $1 < 2$, so $1 \leq 2$ is also true.
On the other hand, we have $1 \leq 1$, but it's not true that $1 < 1$.
Also, you proved $n! < n^n$ for $n > 1$, but this is not true for every positive integer, since for $n=1$ we get the equality. But it's true that $n! \leq n^n$ for every positive integer $n$.
