Is there an upper bound on Bell numbers? For some reason my intuition is that $n^n$ might be an upper bound for Bell numbers, but I can't find anything to confirm that.
Sorry if this is a simple question! (it's been a while since my undergrad math days)
 A: Yes, $n^n$ is an upper bound, and in fact a more precise tighter one exists:

http://en.wikipedia.org/wiki/Bell_number#Growth_rate
A: "For some reason my intuition is that $n^n$ might be an upper bound for Bell numbers" - here is a simple way to confirm the intuition.
The Bell number $B_n$ is the number of ways to partition the set $\{1,..,n\}$.
The number of maps from $\{1,..,n\}$ to $\{1,..,n\}$ is precisely $n^n$ (there are $n$ ways to map $1\in\{1,..,n\}$, $n$ ways to map $2$, ... , $n$ ways to map $n$, so altogether this gives $n\cdot n\cdot  ...  \cdot n = n^n$). Now, each such map $f$ defines a partition of $\{1,..,n\}$: i and j are in the same partition element if $f(i)=f(j)$.  
Moreover, for any given partition $P$ of $\{1,..,n\}$ there exists a corresponding map $f_P$. Indeed it suffices to enumerate the elements of the partition and to map each $i\in\{1,..,n\}$ to the number of the partition element to which $i$ belongs.
So the number of the partitions of $\{1,..,n\}$, which is the Bell number $B_n$, is less or equal to the number of maps from $\{1,..,n\}$ to $\{1,..,n\}$, which is $n^n$. 
Remark: the above correspondence between partitions and functions is not one to one. Indeed, two different maps may correspond to the same partition: take $n=2$ and the maps $f_1,f_2$ where $f_1(0)=0, f_1(1)=1$ and $f_2(0)=1, f_2(1)=0$. Both define the same partition.  
