In Sikorski's book "Boolean Algebras" (3rd edition), p. 42, one finds the following theorem:
In order that $\mathfrak{A}$ be a free Boolean algebra with $n$ free generators, it is necessary and sufficient that $\mathfrak{A}$ be the Boolean product of an indexed set of $n$ four-element Boolean algebras.
Suppose that $\mathfrak{A}$ has 1 generator, hence 4 elements.Then $\mathfrak{A}$ is trivially the Boolean product of 1 four-element Boolean algebra.
Suppose next that $\mathfrak{A}$ has 2 generators, hence 16 elements. Then $\mathfrak{A}$ is the Boolean product of 2 four-element Boolean algebras (which indeed has 16 elements).
Consider now the case that $\mathfrak{A}$ has 3 free generators, hence, 256 elements. Then the Boolean product of 3 four-element Boolean algebras has 64 elements (not 256).
What am I missing here?