Induction principle for $n!<n^n$ How can I prove that $n!&lt;n^n$ for every $n&gt;1$ using the induction principle?

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

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

$$(n+1)!=(n+1)\cdot n! \stackrel{\color{red}{n!&lt;n^n}} &lt;(n+1)\cdot n^n\stackrel{\color{red}{n^n&lt;(n+1)^n}} &lt;(n+1)(n+1)^n$$
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\begin{align}
\pars{n + 1}! &amp; = {\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] &amp; =
\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}\ &lt; 1} \atop {\mbox{denominator} &gt; 1}}}}^{\ds{\Large\color{red}{&lt; 1}}}
\\[5mm] &amp; &lt;
\bbx{\large \pars{n + 1}^{n + 1}}
\end{align}