Polynomial bounds? Q1: Is the function $$\lceil{\lg n}\rceil!$$ polynomial bounded?
Q2: Is the function $$\lceil{\lg\lg n}\rceil!$$ polynomially bounded?
$$\lg = \log_2$$
Polynomially bounded: $f(n)$ is polynomially bounded if for some constants $a$ and $n$, $f(n) < cn^a$, for all $n$ greater than $n_0$.

$\lceil{\log n}\rceil! \geq (\log n)! \approx \bigg(\frac{\log n}{e} \bigg)^{\log n} \sqrt{2 \pi \log n}$ 
At the same time a polynomial function is $n^a, \ a \in \mathbb{Z^+}$, hence 
$$
n^a = e^{\log n^a} = (e^a)^{\log n}
$$
where $e^a$ is a constant. Obviously $\frac{\log n}{e} \geq e^a \ n \ \forall  \geq e^{e^{a+1}}$
Hence $\lceil{\log n}\rceil!$ grows superpolynomially
EDIT:  $\exists c_1 \in \mathbb{R}^{+}$ s.t. $c_1 \log \log n \geq \lceil \log \log n \rceil$. Now denote $c_1 \log \log n=v$, hence $n=e^{e^{\frac{v}{c_1}}}, \ n^{a}=e^{ae^{\frac{v}{c_1}}}$. By Stirling's formula $v! \approx (\frac{v}{e})^{v} \sqrt{2 \pi v} > v^v$. By taking logs we get $\log v^v = v \log v < \log n^a = a e^{\frac{v}{c_1}}$. Hence $\lceil \log \log n \rceil!$ is subpolynomial
