Suppose that expression indeed defines a number, call it $x$: $$x = \sqrt{6\sqrt{6\sqrt{6\sqrt{\cdots}}}}$$ Square both sides: $$x^2 = 6\color{blue}{\sqrt{6\sqrt{6\sqrt{\cdots}}}}$$ Notice that the blue part is $x$ again, so: $$x^2 = 6x$$ This equation has two solutions; but one of them is...
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From the comment:
> why $x \
e 0$ ?
Take a look at the sequence: $$\sqrt{6}, \sqrt{6\sqrt{6}},\sqrt{6\sqrt{6\sqrt{6}}}, \ldots$$ This sequence is increasing with first term $\sqrt{6}>0$ so if this converges, it cannot converge to $0$.
To show this increasing sequence converges, you only need that it is bounded. Call the $n$-th term in the sequence above $x_n$ and observe that $x_1 = \sqrt{6} \le 6$. Now if $x_n \le 6$, then by induction also $x_{n+1} = \sqrt{6x_n} \le \sqrt{6 \cdot 6} = 6$. Thanks to Bungo for his comments.
You can also take a look at this similar question.