I have been trying to prove the following.
Suppose $B$ is a finite non-trivial Boolean algebra. Consider the product $B^{B^B}$. Define for $b\in B$, $\pi_b\colon B^B\rightarrow B$ given by $\pi_b(f)=f(b)$ for all $f\in B^B$. Then note that $\pi_b\in B^{B^B}$; so consider the generated Boolean subalgebra $\langle\pi_b\mid b\in B\rangle$, and show that this has cardinality $2^{2^{|B|}}$.
I suppose one could try to establish a bijection between the subalgebra and the set $2^{2^{B}}$, but I cannot see a natural way to define it. Any help would be appreciated!
It is enough to show that the algebra has $2^{|B|}$ atoms.
Now you have $|B|$ projections and each atom has the shape $$\bigwedge_{b\in B}\pi_b^{\sigma(b)},$$ where $\sigma\in\{-1,1\}^B$, and $\pi_b^1=\pi_b$, and $\pi_b^{-1}=\pi_b'$. Since each $\sigma\in\{-1,1\}^B$ gives a different atom, the number of atoms is $|\{-1,1\}^B|=2^{|B|}$.
Edit. Here I justify that, according to the notation above, each $\sigma$ gives rise to a different atom of the Boolean algebra.
Let us fix $\sigma_1 \in \{-1,1\}^B$ and define $f:B\to B$ by $$ f(x) = \begin{cases} 1, &\text{if} &\sigma_1(x)=1,\\ 0, &\text{if} &\sigma_1(x)=-1. \end{cases} $$ It follows that, for each $b \in B$, $$\pi_b^{\sigma_1(b)}(f)=1,$$ because if $\sigma_1(b)=1$, then $f(b)=1$, whence $\pi_b^{\sigma_1(b)}(f)=\pi_b(f)=f(b)=1$;
if $\sigma_1(b)=-1$, then $f(b)=0$, and $\pi_b^{\sigma_1(b)}(f)=\pi_b'(f)=(f(b))'=0'=1$.
Thus, we conclude that $$\left( \bigwedge_{b\in B} \pi_b^{\sigma_1(b)} \right)(f) = 1.$$
Now, if $\sigma_2\neq\sigma_1$, then there exists $b_0\in B$ such that $\sigma_1(b_0) \neq \sigma_2(b_0)$.
Using the same $f$ as above, we see that $$\pi_{b_0}^{\sigma_2(b_0)}(f)=0$$ because if $\sigma_2(b_0)=1$, then $\sigma_1(b_0)=-1$ and $f(b_0)=0$, whence $$\pi_{b_0}^{\sigma_2(b_0)}(f)=\pi_{b_0}(f)=f(b_0)=0;$$ if $\sigma_2(b_0)=-1$, then $\sigma_1(b_0)=1$ and $f(b_0)=1$, whence $$\pi_{b_0}^{\sigma_2(b_0)}(f)=\pi_{b_0}'(f)=(f(b_0))'=1'=0.$$ Hence $$\left( \bigwedge_{b\in B} \pi_b^{\sigma_2(b)} \right)(f) = 0.$$ So if $\sigma_1 \neq \sigma_2$ then we constructed $f:B\to B$ such that $$\bigwedge_{b\in B}\pi_b^{\sigma_1(b)}(f) \neq \bigwedge_{b\in B}\pi_b^{\sigma_2(b)}(f),$$ and therefore we can conclude that $$\bigwedge_{b\in B}\pi_b^{\sigma_1(b)} \neq \bigwedge_{b\in B}\pi_b^{\sigma_2(b)}.$$