\[3=\sqrt{1+2\sqrt{1+3\sqrt{\cdots}}}\]
It is easy to see that \(n(n+2)=n\sqrt{1+(n+1)(n+2)}\).
Let \(f(n)=n(n+2)\). Then we see that:
\begin{align}
f(n) &= n\sqrt{1+f(n+1)} \\
&= n\sqrt{1+(n+1)\sqrt{1+f(n+2)}} \\
&= n\sqrt{1+(n+1)\sqrt{1+(n+2)\sqrt{1+f(n+3)}}} \\
&=\cdots
\end{align}
And so
\[n(n+2)=n\sqrt{1+(n+1)\sqrt{1+(n+2)\sqrt{1+(n+3)\sqrt{\cdots}}}}\]
For \(n=1\) we have:
\[3=\sqrt{1+2\sqrt{1+3\sqrt{\cdots}}}\]

that \(\sqrt{2}\) is irrational. Suppose that \(\sqrt{2}\in\mathbb Q\). So there exists an integer \(k>0\) such that \(k\sqrt{2}\in\mathbb Z\). Let \(n\) be the smallest one with this property and take \(m=n(\sqrt{2}-1)\). We observe that \(m\) has the following properties:
\[
1. \ m∈Z \\
2. \ m>0 \\
3. \ m\sqrt{2}\in\mathbb Z \\
4. \ m<n
\]
and so we come to a contradiction! Q.E.D.

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