Proving $n! < n^n$ by induction for all $n\geq 2$.

Question

I am having trouble simplifying an induction question. The question is:

Let $P(n)$ be the statement that $n! < n^n$ where $n$ is an integer greater than $1$.

My work so far:

Base case $n = 2$

$2! = 2$

$2^2 = 4$

$2 < 4$. Therefore the base case is true

Inductive Hypothesis

$P(k) = k! < k^k$ for some $k > 1$

Inductive step

$P(k+1) = (k+1)! < (k+1)^{k+1}$

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

(so from my understanding and based on previous questions, I want to take $k^k(k+1)$ into $(k+1)^{k+1}$ as next inductive step.

$k^k(k+1) = k^k+1 + k^k$ using distribution.

I'm not sure where to go from here.

Thanks!

Sorry about the poor formating, this is my first post and I am not sure how to format it so the powers look like powers. — Nickknack, Mar 20 '15 at 17:43
You may want to consider writing up your proof with greater clarity; although you now see "how the proof works," I have added an answer that I hope will make it clearer how your proof could flow more clearly. Hope it helps. :) — Daniel W. Farlow, Mar 20 '15 at 18:23

score 2 · Accepted Answer · answered Mar 20 '15 at 17:46

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You don't need to have equality to continue - an inequality will do. All you have to say is that $k^k<(k+1)^k$ for $k>1$.

answered Mar 20 '15 at 17:46

Peter Woolfitt

21,114

score 1 · Answer 2 · answered Mar 20 '15 at 18:24

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$$\frac{n^n}{n!} = \frac{n^{n-1}}{(n-1)!} > \frac{(n-1)^{n-1}}{(n-1)!} = \frac{(n-1)^{n-2}}{(n-2)!} > \dots > \frac{2^2}{2!} = 2 > 1$$

answered Mar 20 '15 at 18:24

MT_

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score 0 · Answer 3 · answered Mar 20 '15 at 17:46

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Suppose $P(n)$ is true for an arbitrary $n \in \Bbb N$. Then $ n1 \lt n^n $. Using this,

The Inductive Step goes:

$$ (n + 1)! = (n + 1) \cdot n! \lt (n + 1) \cdot n^n \lt (n + 1) \cdot (n + 1)^n $$

answered Mar 20 '15 at 17:46

Ishfaaq

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score 0 · Answer 4 · edited Apr 13 '17 at 12:21

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If $n!<n^n$ for $n=k$ i.e., $k^k<k!$

$(k+1)!=(k+1)\cdot k!<(k+1)k^k$

So, it is sufficient to show $(k+1)k^k<(k+1)^{k+1}\iff\left(1+\dfrac1k\right)^k>1$ which holds true for $k=1$

Use I have to show $(1+\frac1n)^n$ is monotonically increasing sequence

edited Apr 13 '17 at 12:21

Community

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answered Mar 20 '15 at 17:46

lab bhattacharjee

274,582

score 0 · Answer 5 · answered Mar 20 '15 at 18:18

Claim: For $n\geq 2$, let $S(n)$ denote the statement $$ S(n) : n! < n^n. $$ Base step ($n=2$): $S(2)$ holds because $2!=2<4=2^2$.

Inductive step: Fix some $k\geq 2$ and suppose that $S(k)$ is true, where $$ S(k) : k! < k^k. $$ We must show that $$ S(k+1) : (k+1)! < (k+1)^{k+1} $$ follows. Starting with the left-hand side of $S(k+1)$, \begin{align} (k+1)! &= (k+1)\cdot k!\tag{by definition}\\[0.5em] &< (k+1)\cdot k^k\tag{by $S(k)$}\\[0.5em] &< (k+1)\cdot(k+1)^k\tag{since $k\geq 2$}\\[0.5em] &\leq (k+1)^{k+1}, \end{align} we see that the right-hand side of $S(k+1)$ follows. This completes the inductive step.

By mathematical induction, $S(n)$ is true for all $n\geq 2$. $\blacksquare$

Proving $n! < n^n$ by induction for all $n\geq 2$.

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