discrete mathematics - prove by induction that $(a^n - b^n)$ is a multiple of $(a - b)$ for $n geq 1$

Okay so I just finished a final in discrete mathematics and I could not figure out how to finish this proof:

"Prove by mathematical induction, that $(a^n - b^n)$ is a multiple of $(a - b)$ when $a$ and $b$ are integers and $n \geq 1$.

$\underline{Base case: n = 1}$

$(a^1 - b^1) = (a - b)x$

$x = 1$

$a - b = a - b$

$\underline{Inductive Hypothesis:}$

$n = k$ for some $k$

$(a^k - b^k) = (a - b)x$

$\underline{Inductive Step:}$

$n = k + 1$

$(a^{k + 1} - b^{k + 1}) = (a - b)x_1$

$(a \cdot a^{k}) - (b \cdot b^{k}) = (a - b) x_1$

$a \cdot [(a - b)x + b^k] + b \cdot [(a - b)x - a^k]= (a-b)x_1$

$(a -b)xa + ab^k + (a - b)xb - ba^k = (a - b)x_1$

$(a -b)xa + (a - b)xb - ab^k + ba^k = (a - b)x_1$

$(a -b)xa + (a - b)xb + ba^k - ab^k = (a - b)x_1$

.....

I don't know if I did it right but I couldn't get any further than this.

Answer

Assumed $a^k-b^k=(a-b)m,m\in\Bbb Z\implies a^k=b^k+(a-b)m$

We have $a^{k+1}-b^{k+1}=a\times a^k-b^{k+1}=a[b^k+(a-b)m]-b^{k+1}\\=am(a-b)+ab^k-b^{k+1}\\=am(a-b)+b^k(a-b)\\=(am+b^k)(a-b)$