proof $\lfloor\log n\rfloor !$ is an exponential function? Can anyone prove that $\lfloor\log n\rfloor !$ is an exponential function? 

I've tried a lot but i didn't find anything relating to the solution except e number that i guess it can help.

 A: $$\log(\lfloor \log n\rfloor!)=\sum_{k=2}^{\lfloor\log n\rfloor}\log k<\int_2^{\lfloor\log n\rfloor}\log(x+1)\,dx\\
=(\lfloor\log n\rfloor+1)(\log(\lfloor\log n\rfloor+1)-1)-3\log3+3.$$
Taking the antilogarithm, you get an $O(n^{\log\log n})$ expression.
A: An elementary disproof
The family of exponential functions is characterized by the following functional equation
$$f(x+y)=f(x)f(y).$$
For our case, this means that we should be able to show that 
$$\lfloor\log(n+m)\rfloor !=\lfloor\log(n)\rfloor !\lfloor\log(m)\rfloor !.$$
Assume for the sake simplicity that the base of our $\log$ is $10$.
For the integers $N<M$ let $n=10^N, m=10^M$. Then
$$\lfloor\log(10^{N}+10^M)\rfloor!<\lfloor\log(2\times 10^{M})\rfloor!=\lfloor\log(2)+ M\rfloor!= M!$$
and
$$\lfloor\log(10^{N})\rfloor!\lfloor\log(10^{M})\rfloor!=(N!)(M!)>M!.$$
So, our function is not exponential.
A: Let $f(n)=  [\log n] !$ for $n\geq 1.$ We have $f(n)  \leq (1+\log n)!=(\log ne)! .$ 
From Stirling's Formula  $x!=(x/e)^x\sqrt {2\pi x}\;(1+d(x))$ where $\lim_{x\to \infty}d(x)=0$ we have $$f(n)\leq ((\log ne)/e)^{\log ne}\sqrt {2 \pi \log ne}\;(1+d(\log ne)).$$ Taking logs we have $$\log f(n)\leq (\log ne)^2 -\log ne +\log \sqrt {2 \pi}+ (1/2)\log \log  ne+\log (1+d(\log ne)).$$  For any $k>0$ we have therefore, $$\lim_{n\to \infty}  (\log f(n))/\log e^{kn}=(\log f(n))/k\log n=0$$ because $(\log ne)^2/n\to 0$ as $n\to \infty.$
So as $n\to \infty,$ the function  $f(n)$ grows more slowly than any increasing exponential function in $n $. 
Footnotes: (i). FYI, for $n\geq 1$ we have $0<d(n)<1+1/6n.$ (ii). It is not hard to show there exists $L>0$ such that $L=\lim_{n\to \infty} (\log n!)-( (n+1/2) \log n -n),$ which suffices for this Q . Proving that $L=\log \sqrt {2\pi}$ is more difficult.
