Computational complexity of Bell numbers
Yes, the $n$th Bell number is exponential in $n$.  Wikipedia lists the following relation:

$$B_{n+1} = \sum_{k=0}^n {n \choose k} B_k.$$

It follows that

$$B_{n+1} \ge B_n + n B_{n-1} \ge n B_{n-1},$$

so

$$B_{n+1} \ge n(n-2)(n-4)\cdots \ge (n/2)^{n/4},$$

which is exponential in $n$.



This fact has little or nothing to do with NP-hardness; it is just unconditionally true, regardless of whether P = NP or not.

If you have a problem for which every algorithm takes $\Omega(B_n)$ time, then the problem is not in P; it requires exponential time.  It might or might not be in NP.

If you have a problem where the obvious algorithm takes $\Theta(B_n)$ time, that proves nothing (maybe there is some other algorithm you haven't thought of yet that is faster).