Proving for all $n ≥ 2,$ the Bell numbers are less than $n^n$

Prove that, for all even integers n ≥ 2, $$(n/2)^{n/2} \le r_n \le n^n$$ where $$r_n$$ are the Bell numbers.

I know the Bell numbers are given by the following recurrence relation:

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

So far I've tried to use an inductive proof, but after the initial step I'm not sure where to go with it. I also tried the reasoning of how the Bell numbers are directly related to the power set on $$n$$ which can only be as large as $$2^n$$, but the Bell numbers grow faster than this. The answer seems obvious to me, but I lack the tools to prove it myself.

$$B_n$$ is the number of equivalence elations on a set with $$n$$ elements. We can take the set to be $$[n]=\{1,\dots,n\}$$. Put each odd number in a separate equivalence class. Then for each of the $$N/2$$ even numbers, we have $$n/2$$ choices fo the class it belongs to, giving $$\left(\frac{n}{2}\right)^{n/2}$$ different equivalence relations.
On the other hand, imagine that we have $$n$$ distinct buckets, and we put each element in one of the buckets. There are $$n^n$$ ways to do this, and every equivalence relation corresponds to one or more such distributions, so there are fewer than $$n^n$$ equivalence relations.