Is there a function that grows faster than exponentially but slower than a factorial? In big-O notation the complexity class $O(2^n)$ is named "exponential". The complexity class $O(n!)$ is named "factorial".

I believe that $f(n) = O(2^n)$ and $g(n) = O(n!)$ means that $\dfrac{f(n)}{g(n)}$ goes to zero in the limit as $n$ goes to infinity.

Is there any known function between the factorial and exponential complexity classes? 

In other words is there any known function $j(n)$  that dominates every function in $O(2^n)$, such that:
$$ (j(n) \neq O(2^n)) \wedge (j(n) = O(n!)) \wedge (n! \neq O(j(n)))
$$
 or, informally, $j(n)$ grows asymptotically strictly faster than $2^n$ but not as fast as $n!$?

Or perhaps it has been proven that no such function can exist?

Note: this may seem like a Computer Science question, but in fact I am attempting to prove that any periodic, convergent power series must have coefficients whose inverses grow asymptotically as fast as $n!$ but not faster. I think I can show they most grow faster than $O(2^n)$, but that does not prove they are in $\Theta(n!)$ unless there is no complexity class between $O(2^n)$ and $O(n!)$.
Hint For exponential you have the growth $$\frac{a_{n+1}}{a_n}=\mbox{constant}$$

For the factorial you have the growth 
$$\frac{b_{n+1}}{b_n}=n$$

Take any function $g(n)$ which grows to infinity slower than $n$ and  set 
$$\frac{c_{n+1}}{c_n}=g(n)$$

For example, $g(n)=\sqrt{n}$  gives the example $\sqrt{n!}$ given by AntonioVargas.

Another interesting example is $g(n)=\ln(n)$ which gives
$$d_n =\prod_{k=2}^n \ln(k)$$
To add to Especially Lime's answer, there are a couple of common kinds of classes of asymptotic behaviour relative to any given class. As $n \overset{\in\mathbb{N}}\to \infty$, we often naturally encounter not just $O(f(n))$ but also $f(n)^{O(1)}$ and $f(n)^{o(1)}$. (For definition of $o(1)$ see Landau notation.) And there are still asymptotic growth rates between $\log\log(n)^{Θ(1)}$ and $\log(n)^{Θ(1)}$, such as $(\log\log(n))^{\log\log\log(n)}$.