We have proved that

AAa,$bin{Z}^{+},(2^a-1)(mod{2}^b-1)=2^a(modb)-1,[8+12]$

Generalize this result for $n$ in place of $2.$ That is$,$ show that

AAn$>1,$AAa,$bin{Z}^{+},(n^a-1)(mod{n}^b-1)=n^a(modb)-1$

Hint:

$2$

Represent the numbers in radix $=n$ using symbols $n{n}^k-1k$#$d_{n-1}(n^k-1{)}_n-=(d_{n-1}{)}^k(n^k-1{)}_n{d}_i=i,0$i and show that $in$ radix $n,n^k-1is$ a string $ofk$# $of{d}_{n-1}\text{'}s.$ That $is,(n^k-1{)}_n-=(d_{n-1}{)}^k.$

Reason $on$ the strings $of(n^k-1{)}_n.$