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 $\frac{n!}{(n+1)^n} \le \frac{1}{n +1}$
 A: 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 $(n+1)^{n-1}\ge n^{n-1}\ge n!$
A: Looks like a trivial proof holds.  $n!$ has $n$ terms, each $\le n$.  So $\frac{n!}{n^n}=\frac{1}{n}\times C\le \frac{1}{n}$, since $C=\frac{2}{n}\times...\frac{n}{n}\le 1$.
A: If you must have an induction proof, prove the more general inequality:

For $n\geq m\geq 1,$ $$\frac{m!}{n^m}\leq\frac{1}n.$$

You can prove this by induction on $m.$
When $m=1,$ you have $$\frac{m!}{n^m}=\frac1n\leq\frac1n.$$
Assume true for $m$, and assume $n\geq m+1.$
Then also $n\geq m,$ so by induction: $$\frac{m!}{n^{m}}\leq \frac1n.\tag1$$ Also, since $n\geq m+1,$ $$\frac{m+1}{n}\leq 1.\tag2$$
Multiply $(1)$ and $(2)$ together (which we can do because all values are positive) and you get:
$$\frac{(m+1)!}{n^{m+1}}=\frac{m!}{n^m}\cdot\frac{m+1}{n}\leq \frac1n.$$

You can do induction on the original statement, but it takes some finagling. It’s not hard - it amounts to:$$\left(\frac{n+1}n\right)^n\geq\frac{n+1}{n}.$$ But the general theorem here is so direct, and its proof captures the intuitive “non-inductive” proof - that the expression is the product of some number of positive terms, the first $\frac1n,$ and the rest are $\leq1.$
