Expand $(log(n))!$

$$ Using Stirling's approximation for $x$:$\sqrt[\phantom{2}]{2x}(\frac{x}{e}{)}^x$

$(log(n))\sqrt[\phantom{2}]{2*log(n)}(\frac{log(n)}{e}{)}^{log(n)}$

Where $(\frac{log(n)}{e}{)}^{log(n)}=(log(n){)}^{log(n)}*e^{-log(n)}$ and $e^{-log(n)}=\frac{1}{n}$

$(\frac{log(n)}{e}{)}^{log(n)}=(log(n){)}^{log(n)}*\frac{1}{n}$

$So,(log(n))!$~~$\sqrt[\phantom{2}]{2*log(n)}(\frac{log(n{)}^{log(n)}}{n})$

now, $\sqrt[\phantom{2}]{2*log(n)}log(n{)}^{log(n)}$

thus.

$(log(n))!$~~$\frac{log(n{)}^{log(n)}}{n}$

I have gotten this far with trying to find the runtime of $($log$($n$))!.$ Any help on finishing this up for me would be greatly appreciated!