Recursive proof that $n^n \geq n!$ So I'm trying to prove, by induction, that
$$ n^n \geq n!, \forall n\geq1$$
Base case:
$$ \text{For } n=1, 1^1 = 1 \geq 1 = 1!$$
Hypothesis:
$$ n^n \geq n!$$
Step:
$$ \text{Trying to prove: } n^{n+1} \geq (n+1)! $$
Now, somewhere around here I get some contradicting things. For example, if I start from the right side I get:
$$ (n+1)! = (n+1)\cdot n! \leq (n+1)\cdot n^n = n\cdot n^n + n^n = n^{n+1} + n^n$$
Based on this I would need $n^{n+1} + n^n$ to be less than or equal to $n^{n+1}$, which is certainly not true. Something similar happens when I go the other way. 
Any ideas what I'm doing wrong here?
Thanks.
 A: You should try to prove that $$(n+1)^{n+1} \ge (n+1)!$$
\begin{align}
(n+1)^{n+1} &= (n+1) (n+1)^n\\
&\ge(n+1)n^n 
\end{align}
Now use induction hypothesis. 
A: You are trying to prove : 
$$(n+1)^{n+1} \geq (n+1)!$$
Not : 
$$n^{n+1} \geq (n+1)!$$
A: You need to show that $(n+1)^{n+1} \geq (n+1)!$, not that $n^{n+1} \geq (n+1)!$.
Here are some similar questions asked before: you can check your work against any of them if you'd like.


*

*Induction Proof $n! < n^n$

*Show that $n!<n^n $ where $n>1$ and is a Positive Integer

*Proof of $\forall n \in \Bbb N$, $n > 2 \implies n! < n^n$
For future reference, Approach0 is an excellent resource to search for similar questions (much better than the SE functionality itself). All the best.
A: $(n+1)! = (n+1)\cdot n! \leq (n+1)\cdot n^n$ by induction hypothesis, and
$(n+1)\cdot n^n\leq (n+1)\cdot (n+1)^n = (n+1)^{n+1}$. Done.
A: As an alternative:
$$
n! = \underbrace{{n(n-1)(n-2)\cdots 3\cdot 2\cdot 1}}_{n\ \text{times}} \\
n^n = \underbrace{n \cdot n\cdot n \dots n}_{n\ \text{times}}
$$
Note that:
$$
{n! \over n^n} = \frac{n(n-1)(n-2)\cdots 3\cdot 2\cdot 1}{n \cdot n\cdot n \dots n} = \\
= \frac{n}{n} \cdot  \frac{n-1}{n} \cdot \frac{n-2}{n} \cdots  \frac{2}{n} \cdot  \frac{1}{n}
$$
Now note that for any $n \ge 2$:
$$
\frac{n!}{n^n} < 1
$$
