Hashing using Horner’s Rule

When hashing a (key, value) pair where the key is a string, I have seen the following hash function in use:

E.g. $c_n + 256c_{n-1}+ 256^2c_{n-2}+...256^{n-1}c_1$, where this represents the string $c_1c_2..c_n$.
I understand the equation above can be written as:
$c_n + 256(c_{n-1} + 256(c_{n-2}+...256c_1))$, could someone explain how the following is valid, when I have to compute mod over the entire sum, how can we take it term by term?

r= 0;
for i = 1 to n do
r := (c[i] + 256*r) mod TableSize

Let's prove by induction that after $i$ iterations,
$$
r = \sum_{j=1}^i 256^{i-j} c[j] \bmod m,
$$
where $m$ is the size of the table.

The base case is $i = 0$, where $r = 0 = 0 \bmod m$. Now suppose that the formula holds after $i-1$ iterations. After $i$ iterations, we have
\begin{align*}
r &= \left[ c[i] + 256 \left(\sum_{j=1}^{i-1} 256^{i-1-j} c[j] \bmod m\right) \right] \bmod m \\
\\ &= \left[ c[i] + \sum_{j=1}^{i-1} 256^{i-j} c[j] \right] \bmod m
\\ &= \sum_{j=1}^i 256^{i-j} c[j] \bmod m,
\end{align*}
using several times the formulas
$$
(a + b) \bmod m = \left[(a \bmod m) + (b \bmod m)\right] \bmod m, \\
(ab) \bmod m = \left[(a \bmod m)(b \bmod m)\right] \bmod m.
$$

For the proofs of these two formulas check: this and this.

StackExchange Computer Science Q#142141, answer score: 9