Help with a proof in modular arithmetic

Let $a,b,n \in Z$ with $n > 0$ and $a \equiv b \mod n$. Also, let $c_0,c_1,\ldots,c_k \in Z$. Show that :

$c_0 + c_1a + \ldots + c_ka^k \equiv c_0 + c_1b + \ldots + c_kb^k \pmod n$.

For the proof I tried :

$a = b + ny$ for some $y \in Z$.

If I multiply from both side $c_1 + \ldots + c_k$ I obtain :

$c_1a + c_2a + \ldots + c_ka = (c_1b + c_2b + \ldots + c_k) (b + ny)$.

However I can't prove that is true when I multiply by both side $a^1 + a^2 + \ldots + a^k$.

Answer

1) Prove that $k*a \equiv k*b \pmod n$ for any integer $k$.

2) Show that by induction that means $a^k \equiv b^k \pmod n$ for any natural $k$.

3) Show that if $a\equiv b\pmod n$ and $a' \equiv b'\pmod n$ that $a+a'\equiv b +b' \pmod n$.

4) Show your result follows by induction and combinition

....

Or. Note that $a^k - b^k = (a-b)(a^{k-1}+a^{k-2}b + .... +ab^{k-2} + b^{k-1})$.

And that $(c_0 + c_1a + ... + c_ka^k) - (c_0 + c_1b + ... + c_kb^k)=$

$c_1(a-b) + c_2(a^2 - b^2) + ...... c_k(a^k - b^k) =$

.... And therefore......