Proof of $\forall n \in \Bbb N$, $n > 2 \implies n! < n^n$

Question

What I've got so far is this:

Base case: n = 3

then $3 *2 * 1 = 6$ and $3^3 = 27$ $\therefore 6 < 27, 3! < 3^3$

So the base case is true.

So if we assume $n! < n^n$ (n > 2)

$(n + 1)! = (n + 1) * n!$

$(n + 1)n! < (n+1)n^n$ # since $n! <n*n$ (can I do this?)

Here is where I am stuck. I know $n^n < (n+1)^n$ where n > 2 I don't know how to prove this step by induction. However if I were to assume this:

$(n+1)n^n < (n+1)(n+1)^n$

$(n+1)n^n < (n+1)^{n+1}$

Then finally, because of all those inequalities,

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

Does this look correct? Can I assume $n^n < (n+1)^n$? If not, how would I prove it, preferably by induction?

The proof itself, unfortunately yes, but the sub-proof for $n^n < (n+1)^n$, no. — chronocturnal, Jul 24 '14 at 06:26
If you really want to, you can show by induction on $n$ that if $a$ is a positive integer then $a^n \lt (a+1)^n$. — André Nicolas, Jul 24 '14 at 06:30

score 5 · Accepted Answer · answered Jul 24 '14 at 06:29

Use the induction method:

First, take $n=3$, $3! = 6$ and $3^3 =27$, $3! < 3^3$.

Second, assume the inequality holds for $n = K$, $K \in \mathbb{N}$, $K>3$, i.e. $K! < K^K$. Then consider $n= K+1$,

$(K+1)! = (K+1) K! < (K+1) K^K < (K+1) (K+1)^K = (K+1)^{K+1} $,

which is $(K+1)! < (K+1)^{K+1}$.

Proved.

score 3 · Answer 2 · answered Jul 24 '14 at 06:27

Via induction it's a bit tiresome but

Base case $n=2$, $2!=2<4=2^2$ is pretty straightforward.

Then multiplying both sides by $n+1$ gives

$$(n+1)!< (n+1)n^n$$

Considering

$$(n+1)n^n < (n+1)(n+1)^n=(n+1)^{n+1}$$

so by induction we are done.

Again, a direct proof is infinitely easier, for $n\ge 2$, so I include it for comparison's sake.

$$\log(n!)=\sum_{k=1}^n\log k< \sum_{k=1}^n\log n =n\log n=\log (n^n)$$

score 1 · Answer 3 · answered Jul 24 '14 at 06:27

We know that $n!<n^n$ for some $n$, via our inductive hypothesis. We want to show that $(n+1)!<(n+1)^{n+1}$. Your first step is good, multiplying both sides of our inductive hypothesis by $n+1$ to get $(n+1)!<(n+1)n^n$. But $n^n<(n+1)^n$ (we can assume this, if not, it is very easy to prove), so $(n+1)(n^n)<(n+1)(n+1)^n$ and we have that $(n+1)!<(n+1)^{n+1}$ as desired.

score 0 · Answer 4 · answered Jul 24 '14 at 06:55

$$(n+1)!$$

$=$ { by definition of the factorial }

$$(n+1)\ n!$$

$<$ { by the recurrence hypothesis }

$$(n+1)\ n^n$$

$<$ { by monotonicity of the $n^{th}$ power }

$$(n+1)(n+1)^n.$$ $=$ { by distributivity of exponentiaition over multiplication } $$(n+1)^{n+1}.$$ And $$3!<3^3.$$

score 0 · Answer 5 · answered Jul 24 '14 at 07:04

As MathFacts suggested, using Stirling approximation for $n!$ could help. Limited to the very first terms, this approximation write $$n!\simeq n^n \sqrt{2 \pi n}e^{-n}$$ So $n! < n^n$ reduces to $$1 \lt\sqrt{2 \pi n}e^{-n}$$ which is obviously true for any $n \gt 2$.

Proof of $\forall n \in \Bbb N$, $n > 2 \implies n! < n^n$

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