Ytukyg

So I'm trying to prove, by induction, that
$$ n^n geq n!, forall ngeq1$$

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:

$$ (n+1)! = (n+1)cdot n! leq (n+1)cdot n^n = ncdot n^n + n^n = n^{n+1} + n^n$$

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.

$(n+1)! = (n+1)cdot n! leq (n+1)cdot n^n$ by induction hypothesis, and
$(n+1)cdot n^nleq (n+1)cdot (n+1)^n = (n+1)^{n+1}$. Done.

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.

You are trying to prove :

$$(n+1)^{n+1} geq (n+1)!$$

Not :

$$n^{n+1} geq (n+1)!$$

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.

As an alternative:
$$
n! = underbrace{{n(n-1)(n-2)cdots 3cdot 2cdot 1}}_{n text{times}} \
n^n = underbrace{n cdot ncdot n dots n}_{n text{times}}
$$

Note that:
$$
{n! over n^n} = frac{n(n-1)(n-2)cdots 3cdot 2cdot 1}{n cdot ncdot n dots n} = \
= frac{n}{n} cdot frac{n-1}{n} cdot frac{n-2}{n} cdots frac{2}{n} cdot frac{1}{n}
$$

Now note that for any $n ge 2$:
$$
frac{n!}{n^n} < 1
$$