How to prove $n ≡ n_0 + n_1 + \dots +n_k \pmod{b-1}$

Question

I am trying to prove this statement, where $n$ has base $b$ representation, which can be understood easily using this example:

In base $10$, mod $9$ of any number can be found by adding up its digits and doing the mod $9$ of that sum.

It's been a while since I've done proofs, and I'm just not sure where to start here. I know that, for example, in base $10$:

$$10 \bmod 9 = 1\quad \text{or}\quad b \bmod b-1 = 1.$$

I believe I can substitute that in somehow, but I'm not sure how to start. Thanks for your help.

Answer 1

Note that $n = \sum_{i=0}^kn_ib^i$ and $b \equiv 1 \operatorname{mod} (b - 1)$. Now use the fact that if $a \equiv b \operatorname{mod} m$ and $c \equiv d \operatorname{mod} m$, then $a + c \equiv b + d \operatorname{mod} m$ and $ac \equiv bd \operatorname{mod} m$; in particular, $a^n \equiv b^n \operatorname{mod} m$.

Answer 2

Hint: $$((b-1)+1)^k = 1 \quad (\operatorname{mod}\ b-1)$$