Infinite factorials

Start by recognizing that $\sqrt{1\sqrt{1\sqrt{1\ldots}}} = 1, \sqrt{2\sqrt{2\sqrt{2\ldots}}} = 2,$ and so forth because $\sqrt{n\sqrt{n\sqrt{n\ldots}}} = n$ . Why? Because....

$\sqrt{n\sqrt{n\sqrt{n\ldots}}}=n^{\frac{1}{2}} \cdot n^{\frac{1}{4}} \cdot n^{\frac{1}{8}} \cdot \ldots=$ $n^{\text{ }\large{\sum\limits_{i=1}^{\infty} \frac{1}{2^i}}}$ = $n^{1}$ .

Therefore, $\sqrt{6n \times \sqrt{1\sqrt{1\sqrt{1\ldots}}} \times \sqrt{2\sqrt{2\sqrt{2\ldots}}} \times \ldots \times \sqrt{n\sqrt{n\sqrt{n\ldots}}} } = \sqrt{6n(n!)}$ .

Because the expression also equals $12n$ , squaring both sides gives $144n^2=6n(n!)$ . Note that $n!=n(n-1)!$ , so $144n^2=6n^2(n-1)!$

Simplifying gives that $(n-1)!=24$ , so by inspection $n=\boxed{5}$ .

let √n√n√.... be 'X' then X = √n√n√.... and X=√nX

nX=X^2 by solving this quadratic we get two values n and 0, neglect 0. similarly the whole function turns to √6n * 1 * 2 * 3 * .. * n=√6n * n!=√6n^2 * (n-1)!=12n on solving we get

6n^2 * (n-1)!=12^2

(n-1)!=24

n=5