Exponentiation property of the modulo operator

Question

On Khan Academy I have found this property:

$$ A^B \; \text{mod} \; C = ( (A \; \text{mod} \; C)^B ) \; \text{mod} \; C $$

with (of course) $A,B,C \in \mathbb{Z}$.

I was wondering how you can prove it.

I thought to write: $ A = C \cdot m + r $ with $ 0 \le r < C $

So that: $ A \; \text{mod} \; C = r $

And

$ A^B = (C \cdot m + r)^B = \sum\limits_{k=0}^B \binom{B}{k} (C \cdot m)^k r^{B-k}$

But now?

Answer 1

Notice that in your binomial formula all terms in $(Cm)^k\equiv 0\pmod C$ except for $k=0$, so you are left with $r^B$.

Since $r=A \pmod C$ you have proved the statement.