Base $n$ divisibility

Consider the number $A = \overline {a_{t}a_{t-1}a_{t-2}…a_{2}a_{1}a_{0}}_n$ . We gonna to prove that if $m$ is a factor of $n-1$ and $m | (a_{0}+a_{1}+…+a_{t-1}+a_{t})$ , then $m$ divides $A$ .

$\begin{aligned} A=&\sum_{i=0}^{t} a_{i}×n^{i}\\ =&a_{0}+a_{1}\times n+…+a_{t-1}\times n^{t-1}+a_{t}\times n^t \\ =&a_{0}+a_{1}+…a_{t-1}+a_{t}\\ &+a_{1}\times (n-1)+…+a_{t-1}\times (n^{t-1}-1)+a_{t}\times (n^t-1) \end{aligned}$ $A$ is the sum of $a_{0}+a_{1}+…a_{t-1}+a_{t}$ and $a_{1}\times (n-1)+…+a_{t-1}\times (n^{t-1}-1)+a_{t}\times (n^t-1)$ . By the condition, $m |(a_{0}+a_{1}+…a_{t-1}+a_{t})$ . For all $n^i-1\space(i=1,2,…,t)$ , it has the factor $n-1$ , because you can factorize $n^i-1=(n-1)(n^{i-1}+…+1)$ . $m$ is the factor of $n-1$ . Hence, we can conclude that $A$ is the multiple of $m$ .