$\def\l{(\log_b n)!} \def\n{n^3} \def\L{\,\large}$
The fact that $\l$ grows faster than $\n$ can be somewhat counterintuitive.
So, here's a way you can make it more sensible.
Think of what would happen if instead of having a factorial, we had an exponential. That is, instead of having $\l$, we had $e^{\L \log_b n}$.
Since exponentials and logarithms are inverses of each other, we can simplify it a bit.
$$e^{\,\large \log_b n} = \left (b^{\,\large \log_b e} \right )^{\,\large \log_b n} = \left (b^{\,\large \log_b n} \right )^{\,\large \log_b e} = n^{\,\large \log_b e}$$
This means that is we use an exponential instead of a factorial, we get a polynomail with degree $\log_b e$. Depending on the value of $b$, this degree could be anything between $(-\infty, \infty)$, and thus, even greater than $3$!
So, by using exponentials, we can surpass $n^3$.
But that is not all. We also know that factorials grow faster than exponentials! That is, $x! = \omega(c^x)$. So, if we have a factorial with the $(\log n)$, we are already growing faster than any polynomial $n^c$!
And thus, $\l$ grows faster than $\n$