How can I simplify $(n\,\mathrm{ mod }\,b^{k+1})\,\mathrm{ mod }\, b^k$?
$a\equiv\bar a\!\pmod{\!kn}\Rightarrow a\equiv\bar a\!\pmod{\! n};\ $ $(a\bmod kn)\bmod n=a\bmod n.\ $ Congruences persist mod factors of the modulus