Prove $\frac{x^a-1}{x-1}$ is an element of the whole number set

Question

If $x$ and $a$ are elements of the whole number set. Prove that $\frac{x^a-1}{x-1}$ is also an element of the whole number set.

Answer 1

$$\frac{x^a-1}{x-1}=x^{a-1}+x^{a-2}+\cdots+1$$

Answer 2

Notice that $x^a-1=(x-1)(1+x+x^2+x^3...x^{a-1})$

If $x$ and $a$ are members of the natural number set, then so is $1+x+x^2+x^3...x^{a-1}$.

Answer 3

If you write $x^a-1$ in base $x$, all the digits are $x-1$.

For example:

$$10^5-1=99999$$ $$10_{(16}^6-1=\mathrm{FFFFFF}_{(16}$$

Note that $x$ in base $x$ is always written $10$.