Verbality is not direct power-closed

Statement
It is possible to have the following:


 * A group $$G$$
 * A verbal subgroup $$H$$ of $$G$$
 * An infinite cardinal $$\alpha$$ (in fact, we can choose $$\alpha$$ to be the countable cardinal)

such that inside the direct power $$G^\alpha$$ (i.e., the $$\alpha$$-fold external direct product of $$G$$ with itself), the group $$H^\alpha$$ is not verbal.

Related facts

 * Verbality is finite direct power-closed

Proof

 * Let $$G$$ be the free group of countable rank with a freely generating set $$g_1,g_2,\dots,g_n,\dots$$.
 * Let $$H$$ be the verbal subgroup of $$G$$ given as the subset of $$G$$ comprising products of squares.
 * Let $$\alpha = \omega$$, the countable cardinal.

In the group $$G^\alpha$$, $$H^\alpha$$ is the subgroup comprising those elements such that every coordinate is a product of squares. However, the number of squares that we use for each coordinate is not bounded. Consider the element of $$H^\alpha$$ given by:

$$(g_1^2,g_1^2g_2^2,g_1^2g_2^2g_3^2,\dots)$$

This element is not in the verbal subgroup comprising the products of squares, because there is no bound on the number of squares used. We would like to establish a stronger claim, namely that there is no set of words for which $$H^\alpha$$ is the verbal subgroup of $$G^\alpha$$.