Here's the equation : $$W_{N}^{a(2^{n-1}b_{n-1})} =\exp(-2\pi i[a_{n-1}....a_{0}]2^{n-1}b_{n-1}/2^{n})...(16)$$ $$ = \exp(-2\pi i[a_{n-1}...a_{0}]b_{n-1}/2)....(17)$$ $$ = \exp(-2\pi i(2^{n-2}a_{n-1}+...+a_{1}+a_{0}/2)b_{n-1})...(18)$$ $$ = \exp(-2\pi ia_{0}b_{n-1}/2)...(19)$$ $$=W_{2}^{a_{0}b_{n-1}}$$
So how did (18) becomes (19)? by that how did the $2^{n-2}a_{n-1}+...+a_{1}$ evaluates to zero? I know that $W_{m}^{m+k}=W_{m}^{k}$ but what's the $m$ in equation (18) so that I could ignore the rest of the $a$ outside of $a_{0}$?