Is $(\sqrt{n})!=Θ(\sqrt{n}^{\sqrt{n}})$?
Stirling's approximation states that
$$
n! \sim \sqrt{2\pi n} (n/e)^n = \Theta(\sqrt{n} (n/e)^n).
$$
Therefore $\sqrt{n}! = \Theta(n^{1/4} (\sqrt{n}/e)^{\sqrt{n}})$. In particular, $\sqrt{n}! = o(\sqrt{n}^{\sqrt{n}})$.

Note also that if all you want is to express $\sqrt{n}!$ in big Theta notation, you can just write $\sqrt{n}! = \Theta(\sqrt{n}!)$.

If $\log f(n) = \Theta(\log g(n))$ then it doesn't follow that $f(n) = \Theta(g(n))$. For example, $\log (a^n) = \Theta(\log (b^n))$ for all $a,b &gt; 1$, but $a^n$ is not $\Theta(b^n)$ (unless $a = b$).