$3=\sqrt{1+2\sqrt{1+3\sqrt{\cdots}}}$
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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