How using induction it can be shown that:
$$B_n\le n! \;\;\;\;\;\;\;\;\;\left( n\in \mathbb N \right)$$
Where $B_n$ is the nth Bell number.
The base case is true, since $$1=B_0\le 0!=1 \;\;\;\;\;\;\;\;\;\text{and}\;\;\;\;\;\;\;\;\; 1=B_1\le 1!=1$$
Assume the relation holds for $n=k$ and consider $n=k+1$:
This is where I cannot continue, I tried to use the relation $B_{k+1}\ge kB_{k-1}+B_k$, but that did not help me.
Hint:
Consider instead the identity (listed on the Wikipedia page):
$$B_{n+1} = \sum_{k=0}^n \binom{n}{k} B_k$$
Then by (strong) induction, we know each $B_k \leq k!$, so:
$$ B_{n+1} = \sum_{k=0}^n \binom{n}{k} B_k \leq \sum_{k=0}^n \binom{n}{k} k! = \sum_{k=0}^n \frac{n!}{(n-k)!} $$
Can you take it from here?
I hope this helps ^_^