Algorithm to decide if $n \le m!$
Here is a different solution than the one I suggested in the comments. Compute the sequence of factorials $1!,2!,3!,4!,\ldots$ until you get a number which is at least $n$. Since $(k+1)!/k! = O(\log (k!))$, you only compute numbers of magnitude $O(n\log n)$ and so of length $(1+o(1))|n|$ (here $|n|$ is the length of $n$ encoded in binary). Having found the least $k$ such that $n \leq k!$, we know that $n \leq m!$ iff $m \geq k$.

Since the sequence of factorials grows so fast, the running time isn't too bad &mdash; it's quasilinear in $|n|$ (at least in the RAM model). I'll let you work that out.