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How can I prove that $n!<n^n$ for every $n>1$ using the induction principle?

If I put $n=2$ I get $2<4$ so I know that $n!<n^n$ is true.

Now I don't know how to prove that $(n+1)!<(n+1)^{n+1}$. I can write $(n+1)!=n!(n+1)$ but then I don't know how to go on.

user557276
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  • FWIW, this can be proved directly without induction. Both sides are a product of $n$ factors. On the right side, each of the $n$ factors is larger than the corresponding factor on the left (save one). – 2'5 9'2 May 10 '18 at 19:25
  • @user557276 Please recall that if the OP is solved you can evaluate to accept an answer among the given, more details here https://meta.stackexchange.com/questions/5234/how-does-accepting-an-answer-work – user May 31 '18 at 20:48
  • Just to link to the node of this very common inquiry. – Lee David Chung Lin Feb 05 '20 at 12:31

7 Answers7

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HINT

$$(n+1)!=(n+1)\cdot n! \stackrel{\color{red}{n!<n^n}} <(n+1)\cdot n^n\stackrel{\color{red}{n^n<(n+1)^n}} <(n+1)(n+1)^n$$

user
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Hint: $$(n+1)^{n+1} = (n+1)\cdot (n+1)^{n} > (n+1)\cdot n^n$$

Math Lover
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The key is using the induction assumption $n!\lt n^n$

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

Now use $n^n\lt (n+1)^n$ to get

$$(n+1)!\lt (n+1)^{n+1}$$

marwalix
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Base case:

$$1<n\implies 1!<n^1.$$

Inductive step:

For $k<n$,

$$k!<n^k\implies (k+1)!=(k+1)k!<(k+1)n^n\le n^{k+1}.$$

Then for $k=n$,

$$n!<n^n.$$

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But, I don't think the problem needs to apply mathematical induction, because the inequality is pretty obvious.

Since $1<n,2<n,3<n,\cdots,n-1<n,n \leq n$ for $n>1$, by multiplying all of these, we have that $$1 \times 2 \times \cdots \times (n-1) \times n <\underbrace{n \times n \times \cdots \times n}_{\mbox{n factors }}.$$ Thus, $$n!<n^n.$$

WuKong
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Suppose $n>1$ and that $n!<n^n$ holds. Then you can surely write $$ (n+1)!=(n+1)n!<(n+1)n^n $$ You would be done if you can prove that $$ (n+1)n^n\le(n+1)^{n+1} $$ which is equivalent to $$ n^n\le(n+1)^n $$ which in turn is equivalent to $n\le n+1$, which is obviously true.

egreg
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$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} \pars{n + 1}! & = {\pars{n + 1}! \over \pars{n + 1}^{n + 1}} \,\color{blue}{\pars{n + 1}^{n + 1}} = \color{blue}{\pars{n + 1}^{n + 1}}\,{n! \over \pars{n + 1}^{n}} = \color{blue}{\pars{n + 1}^{n + 1}}\,{n\pars{n - 1}\cdots 2 \times 1 \over n^{n}\pars{1 + 1/n}^{n}} \\[5mm] & = \color{blue}{\pars{n + 1}^{n + 1}}\,\ \overbrace{\underbrace{{\pars{1 - 1/n}\pars{1 - 2/n}\cdots \bracks{1 - \pars{n - 2}/n}\bracks{1 - \pars{n - 1}/n} \over \pars{1 + 1/n}^{n}}}_{\ds{{\mbox{numerator}\ < 1} \atop {\mbox{denominator} > 1}}}}^{\ds{\Large\color{red}{< 1}}} \\[5mm] & < \bbx{\large \pars{n + 1}^{n + 1}} \end{align}

Felix Marin
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