Why is $a^n - b^n$ divisible by $a-b$? I did some mathematical induction problems on divisibility


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*$9^n$ $-$ $2^n$ is divisible by 7.

*$4^n$ $-$ $1$ is divisible by 3.

*$9^n$ $-$ $4^n$ is divisible by 5.


Can these be generalized as
$a^n$ $-$ $b^n$$ = (a-b)N$, where N is an integer?
But why is $a^n$ $-$ $b^n$$ = (a-b)N$ ?
I also see that $6^n$ $- 5n + 4$ is divisible by $5$ which is $6-5+4$ and $7^n$$+3n + 8$ is divisible by $9$ which is $7+3+8=18=9\cdot2$.
Are they just a coincidence or is there a theory behind?
Is it about modular arithmetic?

Another proof: 
Denote $r=b/a$. We know that the sum of a geometric progression of the type $1+r+r^2+\ldots+r^{n-1}$ is equal to $\frac{1-r^n}{1-r}$. Thus, we have 
\begin{align}
1-r^n&=(1-r)(1+r+r^2+\ldots+r^{n-1}),\quad\text{substituting $r=b/a$ gives:}\\
a^n-b^n &= (a-b)\color{red}{(a^{n-1}+a^{n-2}b+\ldots+b^{n-1})}\\
a^n-b^n &= (a-b)N
\end{align}
The last step follows since $a,b$ are integers and a polynomial expression of the type in $\color{red}{red}$ font is also an integer.
