<s>[INST] Proving by induction that $\frac{1}{n} \ge \frac{n!}{n^n}$ I have been trying to prove by induction that for all $n \ge 0, \frac{1}{n} \ge \frac{n!}{n^n} $
Here is my proof so far:
Prove that for $ n = 1, \frac{1}{n}  \ge \frac{n!}{n^n} $
$ \frac{1}{1} = 1 $
$ \frac{1!}{1^1} = 1 \le 1 $
Next,  assume there exists $ n \ge 0 $ such that $  \frac{n!}{n^n} \le \frac{1}{n}. $.
We'll prove that $ \frac{1}{n+1} \ge  $ $\frac{(n+1)!}{(n+1)^{n+1}} $.
$\frac{(n+1)!}{(n+1)^{n+1}} = \frac{(n!)(n+1)}{(n+1)^{n+1}} = \frac{n!}{(n+1)^n} \lt \frac{n!}{n^n} \le \frac{1}{n}$
However, I don't know how to show now that $\frac{n!}{(n+1)^n} \le \frac{1}{n +1}$
 [/INST] Just note that $$\frac{1}{n}\ge \frac{n!}{n^n} \iff n^n\ge n! \cdot n$$
$$\iff n^{n-1}\ge n!$$
Now assume that for some $n\ge 1$ the above proposition holds
$$(n+1)^n\ge (n+1)!$$
$$\iff (n+1)^{n-1}\ge n!$$
But this is trivial since $(n+1)^{n-1}\ge n^{n-1}\ge n!$
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