Power congruences in a uniform pro-p group ($x^{p^a}\equiv y^{p^a}(\mathrm{mod}{G}^{p^{a+b}})$)

For a uniform pro-p group with $U_i=G^{p^i}$, the map $x\mapsto x^{p^a}$ transports congruences byCommutator collection gives $(yu{)}^p\equiv y^p{u}^p(\mathrm{mod}{U}_{b+2})$ for $u\in U_b$, and the power map is an isomorphism $U_b/U_{b+1}\to U_{b+1}/U_{b+2}$. Thus a nontrivial first difference moves down exactly one layer when taking a $p$th power. Iterating gives the equivalence; completeness supplies unique roots when the congruence is read in reverse.