Infinite roots factorials

Since $x = \sqrt {x!\sqrt{x!...}}$

We can say that

$\sqrt{x! \times x} = x$

Squaring both sides gives

$x! x = x^{2}$

$x^{2} - x! x = 0$

$x(x - x!) = 0$

Either $x = 0$ (1) or $x - x! = 0$ (2).

In case (1) however, the question specifies that the answer must be a positive integer, and $0$ is neither a positive integer nor a negative integer.

In case (2), we have to solve for $x$ in the following equation:

$x = x!$

The only possible values of $x$ are $1, 2$ since a factorial is the product of all positive integers less than or equal to the number, and in this question, only $x=1$ or $x=2$ works, since $1! = 1$ and $2! = 2 \times 1 = 2$ . If $x$ is greater than $2$ , then $x!$ will be multiplied by $x$ and some other number/numbers which doesn't equal to $1$ , which in turn won't be equal to $x$ .

Therefore only $x = 1$ and $x=2$ satisfy the equation and conditions above, so the answer is $2$ .

Only $1, 2$ satisfy the equation, so the answer is $2$ .

Squaring both sides of this equation yields:

$x! \cdot \sqrt{{x!}\sqrt{x!\sqrt{x!}}....} =x^2$ ;

or $x! \cdot x = x^2;$

or $x^2 \cdot (x-1)! = x^2;$

or $(x-1)! = 1$ ;

or $\boxed{x = 1, 2}.$

Hence, there are two such solutions for $x \in \mathbb{N}.$